Preparation for the specialized level of a single state exam mathematics. Useful materials on trigonometry, large theoretical video lectures, video analysis of problems and a selection of assignments from previous years.
Useful materials
Video collections and online courses
Trigonometric formulas
Geometric illustration of trigonometric formulas
Arc functions. The simplest trigonometric equations
Trigonometric equations
- Necessary theory for solving problems.
- a) Solve the equation $7\cos^2 x - \cos x - 8 = 0$.
b) Find all the roots of this equation, belonging to the interval$\left[ -\dfrac(7\pi)(2); -\dfrac(3\pi)(2) \right]$. - a) Solve the equation $\dfrac(6)(\cos^2 x) - \dfrac(7)(\cos x) + 1 = 0$.
b) Find all the roots of this equation belonging to the interval $\left[ -3\pi; -\pi \right]$. - Solve the equation $\sin\sqrt(16 - x^2) = \dfrac12$.
- a) Solve the equation $2\cos 2x - 12\cos x + 7 = 0$.
b) Find all the roots of this equation belonging to the interval $\left[ -\pi; \dfrac(5\pi)(2) \right]$. - a) Solve the equation $\dfrac(5)(\mathrm(tg)^2 x) - \dfrac(19)(\sin x) + 17 = 0$.
- Solve the equation $\dfrac(2\cos^3 x + 3 \cos^2 x + \cos x)(\sqrt(\mathrm(ctg)x)) = 0$.
- Solve the equation $\dfrac(\mathrm(tg)^3x - \mathrm(tg)x)(\sqrt(-\sin x)) = 0$.
b) Find all the roots of this equation belonging to the interval $\left[ -\dfrac(5\pi)(2); -\pi \right)$.- a) Solve the equation $\cos 2x = \sin\left(\dfrac(3\pi)(2) - x\right)$.
b) Find all the roots of this equation belonging to the interval $\left[ \dfrac(3\pi)(2); \dfrac(5\pi)(2) \right]$. - a) Solve the equation $2\sin^2\left(\dfrac(3\pi)(2) + x\right) = \sqrt3\cos x$.
b) Find all the roots of this equation belonging to the interval $\left[ -\dfrac(7\pi)(2); -2\pi \right]$.
Video analysis of tasks
b) Find all the roots of this equation belonging to the segment $\left[ \sqrt(3); \sqrt(20) \right]$.
b) Find all the roots of this equation belonging to the segment $\left[ -\dfrac(9\pi)(2); -3\pi \right]$.
b) Find all the roots of this equation belonging to the segment $\left[ -\sqrt(3); \sqrt(30) \right]$.
a) Solve the equation $\cos 2x = 1 - \cos\left(\dfrac(\pi)(2) - x\right)$.
b) Find all the roots of this equation belonging to the interval $\left[ -\dfrac(5\pi)(2); -\pi \right)$.
a) Solve the equation $\cos^2 (\pi - x) - \sin \left(x + \dfrac(3\pi)(2) \right) = 0$.
b) Find all the roots of this equation belonging to the interval $\left[\dfrac(5\pi)(2); 4\pi \right]$.
b) Find all the roots of this equation belonging to the interval $\left[\log_5 2; \log_5 20 \right]$.
a) Solve the equation $8 \sin^2 x + 2\sqrt(3) \cos \left(\dfrac(3\pi)(2) - x\right) = 9$.
b) Find all roots of this equation belonging to the interval $\left[- \dfrac(5\pi)(2); -\pi \right]$.
a) Solve the equation $2\log_3^2 (2 \cos x) - 5\log_3 (2 \cos x) + 2 = 0$.
b) Find all the roots of this equation belonging to the interval $\left[\pi; \dfrac(5\pi)(2) \right]$.
a) Solve the equation $\left(\dfrac(1)(49) \right)^(\sin x) = 7^(2 \sin 2x)$.
b) Find all the roots of this equation belonging to the interval $\left[\dfrac(3\pi)(2); 3\pi \right]$.
a) Solve the equation $\sin x + \left(\cos \dfrac(x)(2) - \sin \dfrac(x)(2)\right)\left(\cos \dfrac(x)(2) + \sin \dfrac(x)(2)\right) = 0$.
b) Find all the roots of this equation belonging to the interval $\left[\pi; \dfrac(5\pi)(2)\right]$.
a) Solve the equation $\log_4 (\sin x + \sin 2x + 16) = 2$.
b) Find all the roots of this equation belonging to the interval $\left[ -4\pi; -\dfrac(5\pi)(2) \right]$.
A selection of assignments from previous years
- a) Solve the equation $\dfrac(\sin x)(\sin^2\dfrac(x)(2)) = 4\cos^2\dfrac(x)(2)$.
b) Find all the roots of this equation belonging to the segment $\left[ -\dfrac(9\pi)(2); -3\pi \right]$. (Unified State Exam 2018. Early wave) - a) Solve the equation $\sqrt(x^3 - 4x^2 - 10x + 29) = 3 - x$.
b) Find all the roots of this equation belonging to the segment $\left[ -\sqrt(3); \sqrt(30) \right]$. (USE 2018. Early wave, reserve day) - a) Solve the equation $2 \sin^2 x + \sqrt2 \sin \left(x + \dfrac(\pi)(4)\right) = \cos x $.
b) Find all the roots of this equation belonging to the segment $\left[ -2\pi; -\dfrac(\pi)(2) \right]$. (USE-2018. Main wave) - a) Solve the equation $\sqrt6 \sin^2 x + \cos x = 2\sin\left(x + \dfrac(\pi)(6) \right)$.
b) Find all the roots of this equation belonging to the segment $\left[ 3\pi; \dfrac(9\pi)(2) \right]$. (USE-2018. Main wave) - a) Solve the equation $\sin x + 2\sin\left(2x + \dfrac(\pi)(6) \right) = \sqrt3 \sin 2x + 1$.
b) Find all the roots of this equation belonging to the segment $\left[ -\dfrac(7\pi)(2); -2\pi \right]$. (USE-2018. Main wave) - a) Solve the equation $\cos^2 x + \sin x = \sqrt2 \sin\left(x + \dfrac(\pi)(4) \right)$.
b) Find all the roots of this equation belonging to the segment $\left[ -4\pi; -\dfrac(5\pi)(2) \right]$. (USE-2018. Main wave) - a) Solve the equation $2 \sin\left(2x + \dfrac(\pi)(3) \right) - \sqrt(3) \sin x = \sin 2x + \sqrt3$.
- a) Solve the equation $2\sqrt3 \sin\left(x + \dfrac(\pi)(3) \right) - \cos 2x = 3\cos x - 1$.
b) Find all the roots of this equation belonging to the segment $\left[ 2\pi; \dfrac(7\pi)(2) \right]$. (USE-2018. Main wave) - a) Solve the equation $2\sin\left(2x + \dfrac(\pi)(6) \right) - \cos x = \sqrt3\sin 2x - 1$.
b) Find all the roots of this equation belonging to the segment $\left[ \dfrac(5\pi)(2); 4\pi \right]$. (USE-2018. Main wave) - a) Solve the equation $\sqrt2\sin\left(\dfrac(\pi)(4) + x \right) + \cos 2x = \sin x - 1$.
b) Find all the roots of this equation belonging to the segment $\left[ \dfrac(7\pi)(2); 5\pi \right]$. (USE-2018. Main wave) - a) Solve the equation $\sqrt2\sin\left(2x + \dfrac(\pi)(4) \right) + \sqrt2\cos x = \sin 2x - 1$.
b) Find all the roots of this equation belonging to the segment $\left[ -\dfrac(5\pi)(2); -\pi \right]$. (USE-2018. Main wave) - a) Solve the equation $2\sin\left(x + \dfrac(\pi)(3) \right) + \cos 2x = \sqrt3\cos x + 1$.
b) Find all the roots of this equation belonging to the segment $\left[ -3\pi; -\dfrac(3\pi)(2) \right]$. (USE-2018. Main wave)
b) Find all the roots of this equation belonging to the segment $\left[ \pi; \dfrac(5\pi)(2) \right]$. (USE-2018. Main wave)- a) Solve the equation $2\sin\left(x + \dfrac(\pi)(4) \right) + \cos 2x = \sqrt2\cos x + 1$.
b) Find all the roots of this equation belonging to the segment $\left[ \pi; \dfrac(5\pi)(2) \right]$. (USE-2018. Main wave, reserve day) - a) Solve the equation $2\cos x - \sqrt3 \sin^2 x = 2\cos^3 x$.
b) Find all the roots of this equation belonging to the segment $\left[ -\dfrac(7\pi)(2); -2\pi \right]$. (USE-2018. Main wave, reserve day) - a) Solve the equation $2\cos x + \sin^2 x = 2\cos^3 x$.
b) Find all the roots of this equation belonging to the segment $\left[ -\dfrac(9\pi)(2); -3\pi \right]$. (USE-2018. Main wave, reserve day) - a) Solve the equation $2\sqrt2\sin \left(x + \dfrac(\pi)(3)\right) + 2\cos^2 x = 2 + \sqrt6 \cos x$.
b) Find all the roots of this equation belonging to the segment $\left[ -3\pi; -\dfrac(3\pi)(2) \right]$. (USE-2018. Main wave, reserve day) - a) Solve the equation $x - 3\sqrt(x - 1) + 1 = 0$.
b) Find all the roots of this equation belonging to the segment $\left[ \sqrt(3); \sqrt(20) \right]$. (USE-2018. Main wave, reserve day) - a) Solve the equation $2x\cos x - 8\cos x + x - 4 = 0$.
b) Indicate the roots of this equation that belong to the segment $\left[ -\dfrac(\pi)(2);\ \pi \right]$. (USE 2017, main wave, reserve day) - a) Solve the equation $\log_3 (x^2 - 2x) = 1$.
b) Indicate the roots of this equation that belong to the segment $\left[ \log_2 0(,)2;\ \log_2 5 \right]$. (USE 2017, main wave, reserve day) - a) Solve the equation $\log_3 (x^2 - 24x) = 4$.
b) Indicate the roots of this equation that belong to the segment $\left[ \log_2 0(,)1;\ 12\sqrt(5) \right]$. (USE 2017, main wave, reserve day) - a) Solve the equation $0(,)4^(\sin x) + 2(,)5^(\sin x) = 2$.
b) Indicate the roots of this equation that belong to the segment $\left[ 2\pi;\ \dfrac(7\pi)(2) \right]$. (USE-2017, main wave) - a) Solve the equation $\log_8 \left(7\sqrt(3) \sin x - \cos 2x - 10\right) = 0$.
b) Indicate the roots of this equation that belong to the segment $\left[ \dfrac(3\pi)(2);\ 3\pi \right]$. (USE-2017, main wave) - a) Solve the equation $\log_4 \left(2^(2x) - \sqrt(3) \cos x - 6\sin^2 x\right) = x$.
b) Indicate the roots of this equation that belong to the segment $\left[ \dfrac(5\pi)(2);\ 4\pi \right]$. (USE-2017, main wave) - a) Solve the equation $2\log_2^2 \left(\sin x\right) - 5 \log_2 \left(\sin x\right) - 3 = 0$.
b) Indicate the roots of this equation that belong to the segment $\left[ - 3\pi;\ - \dfrac(3\pi)(2) \right]$. (USE-2017, main wave) - a) Solve the equation $81^(\cos x) - 12\cdot 9^(\cos x) + 27 = 0$.
b) Indicate the roots of this equation that belong to the segment $\left[ - 4\pi;\ - \dfrac(5\pi)(2) \right]$. (USE-2017, main wave) - a) Solve the equation $8^x - 9 \cdot 2^(x + 1) + 2^(5 - x) = 0$.
b) Indicate the roots of this equation that belong to the segment $\left[ \log_5 2;\ \log_5 20 \right]$. (USE 2017, early wave) - a) Solve the equation $2\log^2_9 x - 3 \log_9 x + 1 = 0$.
b) Indicate the roots of this equation that belong to the segment $\left[ \sqrt(10);\ \sqrt(99) \right]$. (USE 2016, main wave, reserve day) - a) Solve the equation $6\log^2_8 x - 5 \log_8 x + 1 = 0$.
b) Indicate the roots of this equation that belong to the segment $\left[ 2;\ 2(,)5 \right]$. (USE 2016, main wave, reserve day) - a) Solve the equation $\sin 2x = 2\sin x + \sin \left(x + \dfrac(3\pi)(2) \right) + 1$.
b) Indicate the roots of this equation that belong to the segment $\left[ -4\pi;\ -\dfrac(5\pi)(2) \right]$. (USE 2016, main wave, reserve day) - a) Solve the equation $2\cos^2 x + 1 = 2\sqrt(2) \cos \left(\dfrac(3\pi)(2) - x \right)$.
b) Indicate the roots of this equation that belong to the segment $\left[ \dfrac(3\pi)(2);\ 3\pi \right]$. (USE-2016, main wave) - a) Solve the equation $2\log^2_2 (2\cos x) - 9 \log_2 (2\cos x) + 4 = 0$.
b) Indicate the roots of this equation that belong to the segment $\left[ -2\pi;\ -\dfrac(\pi)(2) \right]$. (USE-2016, main wave) - a) Solve the equation $8^x - 7 \cdot 4^x - 2^(x + 4) + 112 = 0$.
b) Indicate the roots of this equation that belong to the segment $\left[ \log_2 5;\ \log_2 11 \right]$. (Unified State Exam 2016, early wave) - a) Solve the equation $\cos 2x + \cos^2 \left(\dfrac(3\pi)(2) - x \right) = 0.25$.
b) Indicate the roots of this equation that belong to the segment $\left[ -4\pi;\ -\dfrac(5\pi)(2) \right]$. (Unified State Exam 2016, early wave) - a) Solve the equation $\dfrac(13\sin^2 x - 5\sin x)(13\cos x + 12) = 0$.
b) Indicate the roots of this equation that belong to the segment $\left[ -3\pi;\ -\dfrac(3\pi)(2) \right]$. (Unified State Exam 2016, early wave) - a) Solve the equation $\dfrac(\sin2x)(\sin\left(\dfrac(7\pi)(2) - x \right)) = \sqrt(2)$.
b) Indicate the roots of this equation that belong to the segment $\left$. (USE-2015, main wave) - a) Solve the equation $4 \sin^2 x = \mathrm(tg) x$.
b) Indicate the roots of this equation that belong to the segment $\left[ - \pi;\ 0\right]$. (USE-2015, main wave) - a) Solve the equation $3\cos 2x - 5\sin x + 1 = 0$.
b) Indicate the roots of this equation that belong to the segment $\left[ \pi;\ \dfrac(5\pi)(2)\right]$. (USE-2015, main wave) - a) Solve the equation $\cos 2x - 5\sqrt(2)\cos x - 5 = 0$.
b) Indicate the roots of this equation that belong to the segment $\left[ -3\pi;\ -\dfrac(3\pi)(2)\right]$. (USE-2015, main wave) - a) Solve the equation $\sin 2x + \sqrt(2) \sin x = 2\cos x + \sqrt(2)$.
b) Indicate the roots of this equation that belong to the segment $\left[ \pi;\ \dfrac(5\pi)(2)\right]$. (Unified State Exam 2015, early wave) - a) Solve the equation $2\cos^3 x - \cos^2 x + 2\cos x - 1 = 0$.
b) Indicate the roots of this equation that belong to the segment $\left[ 2\pi;\ \dfrac(7\pi)(2)\right]$. (Unified State Exam 2015, early wave) - a) Solve the equation $\mathrm(tg)^2 x + (1 + \sqrt(3)) \mathrm(tg) x + \sqrt(3) = 0$.
b) Indicate the roots of this equation that belong to the segment $\left[ \dfrac(5\pi)(2); \4\pi\right]$. (USE-2014, main wave) - a) Solve the equation $2\sqrt(3) \cos^2\left(\dfrac(3\pi)(2) + x\right) - \sin 2x = 0$.
b) Indicate the roots of this equation that belong to the segment $\left[ \dfrac(3\pi)(2); \3\pi\right]$. (USE-2014, main wave) - a) Solve the equation $\cos 2x + \sqrt(2) \sin\left(\dfrac(\pi)(2) + x\right) + 1 = 0$.
b) Indicate the roots of this equation that belong to the segment $\left[ -3\pi; \ -\dfrac(3\pi)(2)\right]$. (USE-2014, main wave) - a) Solve the equation $-\sqrt(2) \sin\left(-\dfrac(5\pi)(2) + x\right) \cdot \sin x = \cos x$.
b) Indicate the roots of this equation that belong to the segment $\left[ \dfrac(9\pi)(2); \6\pi\right]$. (Unified State Exam 2014, early wave) - a) Solve the equation $\sin 2x = \sin\left(\dfrac(\pi)(2) + x\right)$.
b) Indicate the roots of this equation belonging to the segment $\left[ -\dfrac(7\pi)(2); \ -\dfrac(5\pi)(2)\right]$. (USE-2013, main wave) - a) Solve the equation $6\sin^2 x + 5\sin\left(\dfrac(\pi)(2) - x\right) - 2 = 0$.
b) Indicate the roots of this equation that belong to the segment $\left[ -5\pi; \ - \dfrac(7\pi)(2)\right]$. (USE-2012, second wave)
The purpose of the lesson:
A) strengthen the ability to solve simple trigonometric equations;
b) teach how to select roots of trigonometric equations from a given interval
During the classes.
1. Updating knowledge.
a)Checking homework: the class is given advanced homework– solve the equation and find a way to select roots from a given interval.
1)cos x= -0.5, where xI [- ]. Answer:.
2) sin x= , where xI . Answer: ; .
3)cos 2 x= -, where xI. Answer:
Students write down the solution on the board, some using a graph, others using the selection method.
At this time class works orally.
Find the meaning of the expression:
a) tg – sin + cos + sin. Answer: 1.
b) 2arccos 0 + 3 arccos 1. Answer: ?
c) arcsin + arcsin. Answer:.
d) 5 arctg (-) – arccos (-). Answer:-.
– Let’s check your homework, open your notebooks with homework.
Some of you found the solution using the selection method, and some using the graph.
2. Conclusion about ways to solve these tasks and statement of the problem, i.e., communication of the topic and purpose of the lesson.
– a) It is difficult to solve using selection if a large interval is given.
– b) Graphic method does not give accurate results, requires verification, and takes a lot of time.
– Therefore, there must be at least one more method, the most universal one - let’s try to find it. So, what are we going to do in class today? (Learn to choose the roots of a trigonometric equation on a given interval.)
– Example 1. (Student goes to the board)
cos x= -0.5, where xI [- ].
Question: What determines the answer to this task? (From general solution equations Let's write the solution in general view). The solution is written on the board
x = + 2?k, where k R.
– Let’s write this solution in the form of a set:
– In your opinion, in what notation of the solution is it convenient to choose roots on the interval? (from the second entry). But this is again a selection method. What do we need to know to get the right answer? (You need to know the values of k).
(Let's make up mathematical model to find k).
since kI Z, then k = 0, hence X= = |
From this inequality it is clear that there are no integer values of k. |
Conclusion: To select roots from a given interval when solving a trigonometric equation, you need to:
- to solve an equation of the form sin x = a, cos x = a It is more convenient to write the roots of the equation as two series of roots.
- to solve equations of the form tan x = a, ctg x = a write down general formula roots.
- create a mathematical model for each solution in the form double inequality and find the integer value of parameter k or n.
- substitute these values into the root formula and calculate them.
Solve example No. 2 and No. 3 from homework using the resulting algorithm. Two students work at the board at the same time, followed by checking the work.
A) Solve equation 2(\sin x-\cos x)=tgx-1.
b) \left[ \frac(3\pi )2;\,3\pi \right].
Show solutionSolution
A) Opening the brackets and moving all the terms into left side, we get the equation 1+2 \sin x-2 \cos x-tg x=0. Considering that \cos x \neq 0, the term 2 \sin x can be replaced by 2 tan x \cos x, we obtain the equation 1+2 tg x \cos x-2 \cos x-tg x=0, which by grouping can be reduced to the form (1-tg x)(1-2 \cos x)=0.
1) 1-tg x=0, tan x=1, x=\frac\pi 4+\pi n, n \in \mathbb Z;
2) 1-2 \cos x=0, \cos x=\frac12, x=\pm \frac\pi 3+2\pi n, n \in \mathbb Z.
b) Using the number circle, select the roots belonging to the interval \left[ \frac(3\pi )2;\, 3\pi \right].
x_1=\frac\pi 4+2\pi =\frac(9\pi )4,
x_2=\frac\pi 3+2\pi =\frac(7\pi )3,
x_3=-\frac\pi 3+2\pi =\frac(5\pi )3.
Answer
A) \frac\pi 4+\pi n, \pm\frac\pi 3+2\pi n, n \in \mathbb Z;
b) \frac(5\pi )3, \frac(7\pi )3, \frac(9\pi )4.
Condition
A) Solve the equation (2\sin ^24x-3\cos 4x)\cdot \sqrt (tgx)=0.
b) Indicate the roots of this equation that belong to the interval \left(0;\,\frac(3\pi )2\right] ;
Show solutionSolution
A) ODZ: \begin(cases) tgx\geqslant 0\\x\neq \frac\pi 2+\pi k,k \in \mathbb Z. \end(cases)
The original equation on the ODZ is equivalent to a set of equations
\left[\!\!\begin(array)(l) 2 \sin ^2 4x-3 \cos 4x=0,\\tg x=0. \end(array)\right.
Let's solve the first equation. To do this we will make a replacement \cos 4x=t, t \in [-1; 1]. Then \sin^24x=1-t^2. We get:
2(1-t^2)-3t=0,
2t^2+3t-2=0,
t_1=\frac12, t_2=-2, t_2\notin [-1; 1].
\cos 4x=\frac12,
4x=\pm\frac\pi 3+2\pi n,
x=\pm \frac\pi (12)+\frac(\pi n)2, n \in \mathbb Z.
Let's solve the second equation.
tg x=0,\, x=\pi k, k \in \mathbb Z.
Using the unit circle, we find solutions that satisfy the ODZ.
The “+” sign marks the 1st and 3rd quarters, in which tg x>0.
We get: x=\pi k, k \in \mathbb Z; x=\frac\pi (12)+\pi n, n \in \mathbb Z; x=\frac(5\pi )(12)+\pi m, m \in \mathbb Z.
b) Let's find the roots belonging to the interval \left(0;\,\frac(3\pi )2\right].
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x=\frac\pi (12), x=\frac(5\pi )(12); x=\pi ; x=\frac(13\pi )(12); x=\frac(17\pi )(12).
Answer
A) \pi k, k \in \mathbb Z; \frac\pi (12)+\pi n, n \in \mathbb Z; \frac(5\pi )(12)+\pi m, m \in \mathbb Z.
b) \pi; \frac\pi (12); \frac(5\pi )(12); \frac(13\pi )(12); \frac(17\pi )(12).
Source: “Mathematics. Preparation for the Unified State Exam 2017. Profile level." Ed. F. F. Lysenko, S. Yu. Kulabukhova.
Condition
A) Solve the equation: \cos ^2x+\cos ^2\frac\pi 6=\cos ^22x+\sin ^2\frac\pi 3;
b) List all roots belonging to the interval \left(\frac(7\pi )2;\,\frac(9\pi )2\right].
Show solutionSolution
A) Because \sin \frac\pi 3=\cos \frac\pi 6, That \sin ^2\frac\pi 3=\cos ^2\frac\pi 6, Means, given equation is equivalent to the equation \cos^2x=\cos ^22x, which, in turn, is equivalent to the equation \cos^2x-\cos ^2 2x=0.
But \cos ^2x-\cos ^22x= (\cos x-\cos 2x)\cdot (\cos x+\cos 2x) And
\cos 2x=2 \cos ^2 x-1, so the equation becomes
(\cos x-(2 \cos ^2 x-1))\,\cdot(\cos x+(2 \cos ^2 x-1))=0,
(2 \cos ^2 x-\cos x-1)\,\cdot (2 \cos ^2 x+\cos x-1)=0.
Then either 2 \cos ^2 x-\cos x-1=0, or 2 \cos ^2 x+\cos x-1=0.
Solving the first equation as quadratic equation relative to \cos x, we get:
(\cos x)_(1,2)=\frac(1\pm\sqrt 9)4=\frac(1\pm3)4. Therefore either \cos x=1 or \cos x=-\frac12. If \cos x=1, then x=2k\pi , k \in \mathbb Z. If \cos x=-\frac12, That x=\pm \frac(2\pi )3+2s\pi , s \in \mathbb Z.
Similarly, solving the second equation, we get either \cos x=-1 or \cos x=\frac12. If \cos x=-1, then the roots x=\pi +2m\pi , m \in \mathbb Z. If \cos x=\frac12, That x=\pm \frac\pi 3+2n\pi , n \in \mathbb Z.
Let's combine the solutions obtained:
x=m\pi , m \in \mathbb Z; x=\pm \frac\pi 3 +s\pi , s \in \mathbb Z.
b) Let's select the roots that fall within a given interval using a number circle.
We get: x_1 =\frac(11\pi )3, x_2=4\pi , x_3 =\frac(13\pi )3.
Answer
A) m\pi, m\in \mathbb Z; \pm \frac\pi 3 +s\pi , s \in \mathbb Z;
b) \frac(11\pi )3, 4\pi , \frac(13\pi )3.
Source: “Mathematics. Preparation for the Unified State Exam 2017. Profile level." Ed. F. F. Lysenko, S. Yu. Kulabukhova.
Condition
A) Solve the equation 10\cos ^2\frac x2=\frac(11+5ctg\left(\dfrac(3\pi )2-x\right) )(1+tgx).
b) Indicate the roots of this equation that belong to the interval \left(-2\pi ; -\frac(3\pi )2\right).
Show solutionSolution
A) 1. According to the reduction formula, ctg\left(\frac(3\pi )2-x\right) =tgx. The domain of definition of the equation will be such values of x such that \cos x \neq 0 and tan x \neq -1. Let's transform the equation using the double angle cosine formula 2 \cos ^2 \frac x2=1+\cos x. We get the equation: 5(1+\cos x) =\frac(11+5tgx)(1+tgx).
notice, that \frac(11+5tgx)(1+tgx)= \frac(5(1+tgx)+6)(1+tgx)= 5+\frac(6)(1+tgx), so the equation becomes: 5+5 \cos x=5 +\frac(6)(1+tgx). From here \cos x =\frac(\dfrac65)(1+tgx), \cos x+\sin x =\frac65.
2. Transform \sin x+\cos x using the reduction formula and the sum of cosines formula: \sin x=\cos \left(\frac\pi 2-x\right), \cos x+\sin x= \cos x+\cos \left(\frac\pi 2-x\right)= 2\cos \frac\pi 4\cos \left(x-\frac\pi 4\right)= \sqrt 2\cos \left(x-\frac\pi 4\right) = \frac65.
From here \cos \left(x-\frac\pi 4\right) =\frac(3\sqrt 2)5. Means, x-\frac\pi 4= arc\cos \frac(3\sqrt 2)5+2\pi k, k \in \mathbb Z,
or x-\frac\pi 4= -arc\cos \frac(3\sqrt 2)5+2\pi t, t \in \mathbb Z.
That's why x=\frac\pi 4+arc\cos \frac(3\sqrt 2)5+2\pi k,k \in \mathbb Z,
or x =\frac\pi 4-arc\cos \frac(3\sqrt 2)5+2\pi t,t \in \mathbb Z.
The found values of x belong to the domain of definition.
b) Let us first find out where the roots of the equation fall at k=0 and t=0. These will be numbers accordingly a=\frac\pi 4+arccos \frac(3\sqrt 2)5 And b=\frac\pi 4-arccos \frac(3\sqrt 2)5.
1. Let us prove the auxiliary inequality:
\frac(\sqrt 2)(2)<\frac{3\sqrt 2}2<1.
Really, \frac(\sqrt 2)(2)=\frac(5\sqrt 2)(10)<\frac{6\sqrt2}{10}=\frac{3\sqrt2}{5}.
Note also that \left(\frac(3\sqrt 2)5\right) ^2=\frac(18)(25)<1^2=1, Means \frac(3\sqrt 2)5<1.
2. From inequalities (1) By the arc cosine property we get:
arccos 1 0 From here \frac\pi 4+0<\frac\pi 4+arc\cos \frac{3\sqrt 2}5<\frac\pi 4+\frac\pi 4,
0<\frac\pi 4+arccos \frac{3\sqrt 2}5<\frac\pi 2,
0 Likewise, -\frac\pi 4 0=\frac\pi 4-\frac\pi 4<\frac\pi 4-arccos \frac{3\sqrt 2}5<
0 For k=-1 and t=-1 we obtain the roots of the equation a-2\pi and b-2\pi. \Bigg(a-2\pi =-\frac74\pi +arccos \frac(3\sqrt 2)5,\, b-2\pi =-\frac74\pi -arccos \frac(3\sqrt 2)5\Bigg). Wherein -2\pi 2\pi For other values of k and t, the roots of the equation do not belong to the given interval. Indeed, if k\geqslant 1 and t\geqslant 1, then the roots are greater than 2\pi. If k\leqslant -2 and t\leqslant -2, then the roots are smaller -\frac(7\pi )2. A) \frac\pi4\pm arccos\frac(3\sqrt2)5+2\pi k, k\in\mathbb Z; b) -\frac(7\pi)4\pm arccos\frac(3\sqrt2)5. Source: “Mathematics. Preparation for the Unified State Exam 2017. Profile level." Ed. F. F. Lysenko, S. Yu. Kulabukhova. A) Solve the equation \sin \left(\frac\pi 2+x\right) =\sin (-2x). b) Find all the roots of this equation that belong to the interval ; A) Let's transform the equation: \cos x =-\sin 2x, \cos x+2 \sin x \cos x=0, \cos x(1+2 \sin x)=0, \cos x=0, x =\frac\pi 2+\pi n, n\in \mathbb Z; 1+2 \sin x=0, \sin x=-\frac12, x=(-1)^(k+1)\cdot \frac\pi 6+\pi k, k \in \mathbb Z. b) We find the roots belonging to the segment using the unit circle. The indicated interval contains a single number \frac\pi 2. A) \frac\pi 2+\pi n, n \in \mathbb Z; (-1)^(k+1)\cdot \frac\pi 6+\pi k, k \in \mathbb Z; b) \frac\pi 2. Source: “Mathematics. Preparation for the Unified State Exam 2017. Profile level." Ed. F. F. Lysenko, S. Yu. Kulabukhova. Means, \sin x \neq 1. Divide both sides of the equation by a factor (\sin x-1), different from zero. We get the equation \frac 1(1+\cos 2x)=\frac 1(1+\cos (\pi +x)), or equation 1+\cos 2x=1+\cos (\pi +x). Applying the reduction formula on the left side and the reduction formula on the right side, we obtain the equation 2 \cos ^2 x=1-\cos x. This equation is by substitution \cos x=t, Where -1 \leqslant t \leqslant 1 reduce it to square: 2t^2+t-1=0, whose roots t_1=-1 And t_2=\frac12. Returning to the variable x, we get \cos x = \frac12 or \cos x=-1, where x=\frac \pi 3+2\pi m, m \in \mathbb Z, x=-\frac \pi 3+2\pi n, n \in \mathbb Z, x=\pi +2\pi k, k \in \mathbb Z. b) Let's solve inequalities 1) -\frac(3\pi )2 \leqslant \frac(\pi )3+2\pi m \leqslant -\frac \pi 2 , 2) -\frac(3\pi )2 \leqslant -\frac \pi 3+2\pi n \leqslant -\frac \pi (2,) 3) -\frac(3\pi )2 \leqslant \pi+2\pi k \leqslant -\frac \pi 2 , 1)
-\frac(3\pi )2 \leqslant \frac(\pi )3+2\pi m \leqslant -\frac \pi 2 , -\frac32\leqslant \frac13+2m \leqslant -\frac12 -\frac(11)6 \leqslant 2m\leqslant -\frac56 , -\frac(11)(12) \leqslant m \leqslant -\frac5(12). \left [-\frac(11)(12);-\frac5(12)\right]. 2)
-\frac (3\pi) 2 \leqslant -\frac(\pi )3+2\pi n \leqslant -\frac(\pi )(2), -\frac32 \leqslant -\frac13 +2n \leqslant -\frac12 , -\frac76 \leqslant 2n \leqslant -\frac1(6), -\frac7(12) \leqslant n \leqslant -\frac1(12). There are no integers in the range \left[ -\frac7(12) ; -\frac1(12)\right]. 3)
-\frac(3\pi )2 \leqslant \pi +2\pi k\leqslant -\frac(\pi )2, -\frac32 \leqslant 1+2k\leqslant -\frac12, -\frac52 \leqslant 2k \leqslant -\frac32, -\frac54 \leqslant k \leqslant -\frac34. This inequality is satisfied by k=-1, then x=-\pi. A) \frac \pi 3+2\pi m; -\frac \pi 3+2\pi n; \pi +2\pi k, m, n, k \in \mathbb Z; b) -\pi . In this article I will try to explain 2 ways selecting roots in a trigonometric equation: using inequalities and using the trigonometric circle. Let's move straight to an illustrative example and we'll figure out how things work. A) Solve the equation sqrt(2)cos^2x=sin(Pi/2+x) Let's solve point a. Let's use the reduction formula for sine sin(Pi/2+x) = cos(x) Sqrt(2)cos^2x = cosx Sqrt(2)cos^2x - cosx = 0 Cosx(sqrt(2)cosx - 1) = 0 X1 = Pi/2 + Pin, n ∈ Z Sqrt(2)cosx - 1 = 0 Cosx = 1/sqrt(2) Cosx = sqrt(2)/2 X2 = arccos(sqrt(2)/2) + 2Pin, n ∈ Z X2 = Pi/4 + 2Pin, n ∈ Z Let's solve point b. 1) Selection of roots using inequalities Here everything is done simply, we substitute the resulting roots into the interval given to us [-7Pi/2; -2Pi], find integer values for n. 7Pi/2 less than or equal to Pi/2 + Pin less than or equal to -2Pi We immediately divide everything by Pi 7/2 less than or equal to 1/2 + n less than or equal to -2 7/2 - 1/2 less than or equal to n less than or equal to -2 - 1/2 4 less than or equal to n less than or equal to -5/2 The integer n in this interval are -4 and -3. This means that the roots belonging to this interval will be Pi/2 + Pi(-4) = -7Pi/2, Pi/2 + Pi(-3) = -5Pi/2 Similarly we make two more inequalities 7Pi/2 less than or equal to Pi/4 + 2Pin less than or equal to -2Pi There are no whole n in this interval 7Pi/2 less than or equal to -Pi/4 + 2Pin less than or equal to -2Pi One integer n in this interval is -1. This means that the selected root on this interval is -Pi/4 + 2Pi*(-1) = -9Pi/4. So the answer in point b: -7Pi/2, -5Pi/2, -9Pi/4 2) Selection of roots using a trigonometric circle To use this method you need to understand how this circle works. I will try to explain in simple language how I understand this. I think in schools, during algebra lessons, this topic was explained many times with clever words from the teacher, in textbooks there were complex formulations. Personally, I understand this as a circle that can be walked around an infinite number of times, this is explained by the fact that the sine and cosine functions are periodic. Let's go around counterclockwise Let's go around 2 times counterclockwise Let's go around 1 time clockwise (the values will be negative) Let's return to our question, we need to select roots in the interval [-7Pi/2; -2Pi] To get to the numbers -7Pi/2 and -2Pi you need to go around the circle counterclockwise twice. In order to find the roots of the equation on this interval, you need to estimate and substitute. Consider x = Pi/2 + Pin. Approximately what should n be for x to be somewhere in this range? We substitute, let's say -2, we get Pi/2 - 2Pi = -3Pi/2, obviously this is not included in our interval, so we take less than -3, Pi/2 - 3Pi = -5Pi/2, this is suitable, let's try again -4 , Pi/2 - 4Pi = -7Pi/2, also suitable. Reasoning similarly for Pi/4 + 2Pin and -Pi/4 + 2Pin, we find another root -9Pi/4. Comparison of two methods. The first method (using inequalities) is much more reliable and much easier to understand, but if you really get serious about the trigonometric circle and the second selection method, then selecting roots will be much faster, you can save about 15 minutes on the exam. Task No. 1 The logic is simple: we will do as we did before, regardless of the fact that now trigonometric functions have a more complex argument! If we were to solve an equation of the form: Then we would write down the following answer: Or (since) But now our role is played by this expression: Then we can write: Our goal with you is to make sure that the left side stands simply, without any “impurities”! Let's gradually get rid of them! First, let’s remove the denominator at: to do this, multiply our equality by: Now let's get rid of it by dividing both parts: Now let's get rid of the eight: The resulting expression can be written as 2 series of solutions (by analogy with a quadratic equation, where we either add or subtract the discriminant) We need to find the largest negative root! It is clear that we need to sort through. Let's look at the first episode first: It is clear that if we take, then as a result we will receive positive numbers, but they do not interest us. So you need to take it negative. Let be. When the root will be narrower: And we need to find the greatest negative!! This means that going in the negative direction no longer makes sense here. And the largest negative root for this series will be equal to. Now let's look at the second series: And again we substitute: , then: Not interested! Then it makes no sense to increase any more! Let's reduce it! Let then: Fits! Let be. Then Then - the largest negative root! Answer: Task No. 2 We solve again, regardless of the complex cosine argument: Now we express again on the left: Multiply both sides by Divide both sides by All that remains is to move it to the right, changing its sign from minus to plus. We again get 2 series of roots, one with and the other with. We need to find the largest negative root. Let's look at the first episode: It is clear that we will get the first negative root at, it will be equal to and will be the largest negative root in 1 series. For the second series The first negative root will also be obtained at and will be equal to. Since, then is the largest negative root of the equation. Answer: . Task No. 3 We solve, regardless of the complex tangent argument. Now, it doesn’t seem complicated, right? As before, we express on the left side: Well, that’s great, there’s only one series of roots here! Let's find the largest negative again. It is clear that it turns out if you put it down. And this root is equal. Answer: Now try to solve the following problems yourself. Ready? Let's check. I will not describe in detail the entire solution algorithm; it seems to me that it has already received enough attention above. Well, is everything right? Oh, those nasty sinuses, there’s always some kind of trouble with them! Well, now you can solve simple trigonometric equations! Let's express The smallest positive root is obtained if we put, since, then Answer: The smallest positive root is obtained at. It will be equal. Answer: . When we get, when we have. Answer: . This knowledge will help you solve many problems that you will encounter in the exam. If you are applying for a “5” rating, then you just need to proceed to reading the article for mid-level which will be devoted to solving more complex trigonometric equations (task C1). In this article I will describe solving more complex trigonometric equations and how to select their roots. Here I will draw on the following topics: More complex trigonometric equations are the basis for advanced problems. They require both solving the equation itself in general form and finding the roots of this equation belonging to a certain given interval. Solving trigonometric equations comes down to two subtasks: It should be noted that the second is not always required, but in most examples selection is still required. But if it is not required, then we can sympathize with you - this means that the equation is quite complex in itself. My experience in analyzing C1 problems shows that they are usually divided into the following categories. To put it simply: if you get caught one of the equations of the first three types, then consider yourself lucky. For them, as a rule, you additionally need to select roots belonging to a certain interval. If you come across a type 4 equation, then you are less lucky: you need to tinker with it longer and more carefully, but quite often it does not require additional selection of roots. Nevertheless, I will analyze this type of equations in the next article, and this one I will devote to solving equations of the first three types. The most important thing you need to remember to solve this type of equation is As practice shows, as a rule, this knowledge is sufficient. Let's look at some examples: Here, as I promised, the reduction formulas work: Then my equation will look like this: Then my equation will take the following form: A short-sighted student might say: now I’ll reduce both sides by, get the simplest equation and enjoy life! And he will be bitterly mistaken! So what to do? Yes, it’s simple, move everything to one side and take out the common factor: Well, we factored it into factors, hurray! Now let's decide: The first equation has roots: And the second: This completes the first part of the problem. Now you need to select the roots: The gap is like this: Or it can also be written like this: Well, let's take the roots: First, let's work with the first episode (and it's simpler, to say the least!) Since our interval is entirely negative, there is no need to take non-negative ones, they will still give non-negative roots. Let's take it, then - it's too much, it doesn't hit. Let it be, then - I didn’t hit it again. One more try - then - yes, I got it! The first root has been found! I shoot again: then I hit again! Well, one more time: : - this is already a flight. So from the first series there are 2 roots belonging to the interval: . We are working with the second series (we are building to the power according to the rule): Undershoot! Missing it again! Missing it again! Got it! Flight! Thus, my interval has the following roots: This is the algorithm we will use to solve all other examples. Let's practice together with one more example. Solution: Again the notorious reduction formulas: Don't try to cut back again! The first equation has roots: And the second: Now again the search for roots. I’ll start with the second episode, I already know everything about it from the previous example! Look and make sure that the roots belonging to the interval are as follows: Now the first episode and it’s simpler: If - suitable If that's fine too If it’s already a flight. Then the roots will be as follows: Well, is the technique clear to you? Does solving trigonometric equations not seem so difficult anymore? Then quickly solve the following problems yourself, and then we will solve other examples: And again the reduction formula: First series of roots: Second series of roots: We begin selection for the gap Answer: , . Quite a tricky grouping into factors (I’ll use the double angle sine formula): then or This is a general solution. Now we need to select the roots. The trouble is that we cannot tell the exact value of an angle whose cosine is equal to one quarter. Therefore, I can’t just get rid of the arc cosine - such a shame! What I can do is figure out that so, so, then. Let's create a table: interval: Well, through painful searches we came to the disappointing conclusion that our equation has one root on the indicated interval: \displaystyle arccos\frac(1)(4)-5\pi A frightening looking equation. However, it can be solved quite simply by applying the double angle sine formula: Let's reduce it by 2: Let's group the first term with the second and the third with the fourth and take out the common factors: It is clear that the first equation has no roots, and now let’s consider the second: In general, I was going to dwell a little later on solving such equations, but since it turned up, there’s nothing to do, I have to solve it... Equations of the form: This equation is solved by dividing both sides by: Thus, our equation has a single series of roots: We need to find those that belong to the interval: . Let's build a table again, as I did earlier: Answer: . Equations reduced to the form: Well, now it’s time to move on to the second portion of equations, especially since I’ve already spilled the beans on what the solution to trigonometric equations of a new type consists of. But it is worth repeating that the equation is of the form Solved by dividing both sides by cosine: Example 1. The first one is quite simple. Move to the right and apply the double angle cosine formula: Yeah! Equation of the form: . I divide both parts by We do root screening: Gap: Answer: Example 2. Everything is also quite trivial: let’s open the brackets on the right: Basic trigonometric identity: Sine of double angle: Finally we get: Root screening: interval. Answer: . Well, how do you like the technique, isn’t it too complicated? I hope not. We can immediately make a reservation: in their pure form, equations that immediately reduce to an equation for the tangent are quite rare. Typically, this transition (division by cosine) is only part of a more complex problem. Here's an example for you to practice: Let's check: The equation can be solved immediately; it is enough to divide both sides by: Root screening: Answer: . One way or another, we have yet to encounter equations of the type that we have just examined. However, it is too early for us to call it a day: there is still one more “layer” of equations that we have not analyzed. So: Everything is transparent here: we look closely at the equation, simplify it as much as possible, make a substitution, solve it, make a reverse substitution! In words everything is very easy. Let's see in action: Example. Well, here the replacement itself suggests itself to us! Then our equation will turn into this: The first equation has roots: And the second one is like this: Now let's find the roots belonging to the interval Answer: . Let's look at a slightly more complex example together: Here the replacement is not immediately visible, moreover, it is not very obvious. Let's first think: what can we do? We can, for example, imagine And at the same time Then my equation will take the form: And now attention, focus: Let's divide both sides of the equation by: Suddenly you and I have a quadratic equation relative! Let's make a replacement, then we get: The equation has the following roots: Unpleasant second series of roots, but nothing can be done! We select roots in the interval. We also need to consider that Since and, then Answer: To reinforce this before you solve the problems yourself, here’s another exercise for you: Here you need to keep your eyes open: we now have denominators that can be zero! Therefore, you need to be especially attentive to the roots! First of all, I need to rearrange the equation so that I can make a suitable substitution. I can’t think of anything better now than to rewrite the tangent in terms of sine and cosine: Now I will move from cosine to sine using the basic trigonometric identity: And finally, I’ll bring everything to a common denominator: Now I can move on to the equation: But at (that is, at). Now everything is ready for replacement: Then or However, note that if, then at the same time! Who suffers from this? The problem with the tangent is that it is not defined when the cosine is equal to zero (division by zero occurs). Thus, the roots of the equation are: Now we sift out the roots in the interval: Thus, our equation has a single root on the interval, and it is equal. You see: the appearance of a denominator (just like the tangent, leads to certain difficulties with the roots! Here you need to be more careful!). Well, you and I have almost finished analyzing trigonometric equations; there is very little left - to solve two problems on your own. Here they are. Decided? Isn't it very difficult? Let's check: Substitute into the equation: Let's rewrite everything through cosines to make it easier to make the replacement: Now it's easy to make a replacement: It is clear that is an extraneous root, since the equation has no solutions. Then: We are looking for the roots we need in the interval Answer: . Then or Answer: Well, that's it now! But solving trigonometric equations does not end there; we are left behind in the most difficult cases: when the equations contain irrationality or various kinds of “complex denominators.” We will look at how to solve such tasks in an article for an advanced level. In addition to the trigonometric equations discussed in the previous two articles, we will consider another class of equations that require even more careful analysis. These trigonometric examples contain either irrationality or a denominator, which makes their analysis more difficult. However, you may well encounter these equations in Part C of the exam paper. However, every cloud has a silver lining: for such equations, as a rule, the question of which of its roots belongs to a given interval is no longer raised. Let's not beat around the bush, but let's go straight to trigonometric examples. Example 1. Solve the equation and find the roots that belong to the segment. Solution: We have a denominator that should not be equal to zero! Then solving this equation is the same as solving the system Let's solve each of the equations: And now the second one: Now let's look at the series: It is clear that this option does not suit us, since in this case our denominator is reset to zero (see the formula for the roots of the second equation) If, then everything is in order, and the denominator is not zero! Then the roots of the equation are as follows: , . Now we select the roots belonging to the interval. Then the roots are as follows: You see, even the appearance of a small disturbance in the form of the denominator significantly affected the solution of the equation: we discarded a series of roots that nullified the denominator. Things can get even more complicated if you come across trigonometric examples that are irrational. Example 2. Solve the equation: Solution: Well, at least you don’t have to take away the roots, and that’s good! Let's first solve the equation, regardless of irrationality: So, is that all? No, alas, it would be too easy! We must remember that only non-negative numbers can appear under the root. Then: The solution to this inequality is: Now it remains to find out whether part of the roots of the first equation inadvertently ended up where the inequality does not hold. To do this, you can again use the table: Thus, one of my roots “fell out”! It turns out if you put it down. Then the answer can be written as follows: Answer: You see, the root requires even more attention! Let's make it more complicated: let now I have a trigonometric function under my root. Example 3. As before: first we will solve each one separately, and then we will think about what we have done. Now the second equation: Now the most difficult thing is to find out whether negative values are obtained under the arithmetic root if we substitute the roots from the first equation there: The number must be understood as radians. Since a radian is approximately degrees, then radians are on the order of degrees. This is the corner of the second quarter. What is the sign of the cosine of the second quarter? Minus. What about sine? Plus. So what can we say about the expression: It's less than zero! This means that it is not the root of the equation. Now it's time. Let's compare this number with zero. Cotangent is a function decreasing in 1 quarter (the smaller the argument, the greater the cotangent). radians are approximately degrees. In the same time since, then, and therefore Answer: . Could it get any more complicated? Please! It will be more difficult if the root is still a trigonometric function, and the second part of the equation is again a trigonometric function. The more trigonometric examples the better, see below: Example 4. The root is not suitable due to the limited cosine Now the second one: At the same time, by definition of a root: We need to remember the unit circle: namely, those quarters where the sine is less than zero. What are these quarters? Third and fourth. Then we will be interested in those solutions of the first equation that lie in the third or fourth quarter. The first series gives roots lying at the intersection of the third and fourth quarters. The second series - diametrically opposite to it - gives rise to roots lying on the border of the first and second quarters. Therefore, this series is not suitable for us. Answer: , And again trigonometric examples with "difficult irrationality". Not only do we have the trigonometric function under the root again, but now it’s also in the denominator! Example 5. Well, nothing can be done - we do as before. Now we work with the denominator: I don’t want to solve the trigonometric inequality, so I’ll do something cunning: I’ll take and substitute my series of roots into the inequality: If - is even, then we have: since all angles of the view lie in the fourth quarter. And again the sacred question: what is the sign of the sine in the fourth quarter? Negative. Then the inequality If -odd, then: In which quarter does the angle lie? This is the corner of the second quarter. Then all the corners are again the corners of the second quarter. The sine there is positive. Just what you need! So the series: Fits! We deal with the second series of roots in the same way: We substitute into our inequality: If - even, then First quarter corners. The sine there is positive, which means the series is suitable. Now if - odd, then: fits too! Well, now we write down the answer! Answer: Well, this was perhaps the most labor-intensive case. Now I offer you problems to solve on your own. Solutions: Second equation: Selection of roots that belong to the interval Answer: Or Let's consider: . If - even, then Answer: , . Or Second part: At the same time, according to DZ it is required that We check the roots found in the first equation: If the sign: First quarter angles where the tangent is positive. Doesn't fit! Fourth quarter corner. There the tangent is negative. Fits. We write down the answer: Answer: , . We have looked at complex trigonometric examples together in this article, but you should solve the equations yourself. A trigonometric equation is an equation in which the unknown is strictly under the sign of the trigonometric function. There are two ways to solve trigonometric equations: The first way is using formulas. The second way is through the trigonometric circle. Allows you to measure angles, find their sines, cosines, etc.Answer
Condition
Solution
Answer
Condition
is not included in the DZ. Answer
b) Find all the roots of this equation belonging to the interval [-7Pi/2; -2Pi]
x3 = -arccos(sqrt(2)/2) + 2Pin, n ∈ Z
x3 = -Pi/4 + 2Pin, n ∈ Z
-15/8 less than or equal to n less than or equal to -9/8
-13/8 less than or equal to n less than or equal to -7/8
Homework or 3 tasks to solve independently.
In the answer to the pi-shi-th-the-smallest-possible root.
In the answer to the pi-shi-th-the-smallest-possible root.Check out the solutions and answers:
Task No. 1
Task No. 2
Task No. 3
AVERAGE LEVEL
Four categories of tasks of increased complexity (formerly C1)
Equations that reduce to factorization
Example 1. Equation reduced to factorization using the reduction and double angle sine formulas
REMEMBER: YOU CAN NEVER REDUCE BOTH SIDES OF A TRIGONOMETRIC EQUATION BY A FUNCTION CONTAINING AN UNKNOWN! SO YOU LOSE YOUR ROOTS!
Example 2. Equation reduced to factorization using reduction formulas
Independent work. 3 equations.
Find all the roots of this equation that lie above the interval.
Indicate the roots of the equation that lie above the cut
Find all the roots of this equation that lie between them.Equation 1.
Equation 2. Checking independent work.
Equation 3: Independent work test.
Indicate the roots of the equation that lie above the cut.
Indicate the roots of the equation that lie between them.Solving trigonometric equations by changing variables
- fits
- overkill
Find all the roots of this equation that lie above the cut.
Indicate the roots of this equation, located above the cut.
Here the replacement is immediately visible:
- fits!
- fits!
- fits!
- fits!
- a lot of!
- also a lot!
ADVANCED LEVEL
- not suitable
- fits
- fits
- fits
overkill
overkill
: , But
No!
Yes!
Yes!
,Training
First equation:
or
ODZ of the root:
or
But
- doesn't fit!
If - odd, : - suitable!
This means that our equation has the following series of roots:
or
Selection of roots in the interval:
- not suitable
- fits
- fits
- a lot of
- fits
a lot of
Since, then the tangent is not defined. We immediately discard this series of roots!
If the sign:SUMMARY AND BASIC FORMULAS