Trigonometric equations are not an easy topic. They are too diverse.) For example, these:

sin 2 x + cos3x = ctg5x

sin(5x+π /4) = cot(2x-π /3)

sinx + cos2x + tg3x = ctg4x

Etc...

But these (and all other) trigonometric monsters have two common and obligatory features. First - you won’t believe it - there are trigonometric functions in the equations.) Second: all expressions with x are found within these same functions. And only there! If X appears somewhere outside, For example, sin2x + 3x = 3, this will already be an equation mixed type. Such equations require individual approach. We will not consider them here.

We will not solve evil equations in this lesson either.) Here we will deal with the simplest trigonometric equations. Why? Yes because the solution any trigonometric equations consists of two stages. At the first stage, the evil equation is reduced to a simple one through a variety of transformations. On the second, this simplest equation is solved. No other way.

So, if you have problems at the second stage, the first stage does not make much sense.)

What do elementary trigonometric equations look like?

sinx = a

cosx = a

tgx = a

ctgx = a

Here A stands for any number. Any.

By the way, inside a function there may not be a pure X, but some kind of expression, like:

cos(3x+π /3) = 1/2

etc. This complicates life, but does not affect the method of solving a trigonometric equation.

How to solve trigonometric equations?

Trigonometric equations can be solved in two ways. The first way: using logic and the trigonometric circle. We will look at this path here. The second way - using memory and formulas - will be discussed in the next lesson.

The first way is clear, reliable, and difficult to forget.) It is good for solving trigonometric equations, inequalities, and all sorts of tricky non-standard examples. Logic is stronger than memory!)

Solving equations using a trigonometric circle.

We include elementary logic and the ability to use the trigonometric circle. Don't you know how? However... You will have a hard time in trigonometry...) But it doesn’t matter. Take a look at the lessons "Trigonometric circle...... What is it?" and "Measuring angles on a trigonometric circle." Everything is simple there. Unlike textbooks...)

Oh, you know!? And even mastered “Practical work with the trigonometric circle”!? Congratulations. This topic will be close and understandable to you.) What is especially pleasing is that the trigonometric circle does not care what equation you solve. Sine, cosine, tangent, cotangent - everything is the same for him. There is only one solution principle.

So we take any elementary trigonometric equation. At least this:

cosx = 0.5

We need to find X. Speaking in human language, you need find the angle (x) whose cosine is 0.5.

How did we previously use the circle? We drew an angle on it. In degrees or radians. And right away saw trigonometric functions of this angle. Now let's do the opposite. Let's draw a cosine on the circle equal to 0.5 and immediately we'll see corner. All that remains is to write down the answer.) Yes, yes!

Draw a circle and mark the cosine equal to 0.5. On the cosine axis, of course. Like this:

Now let's draw the angle that this cosine gives us. Hover your mouse over the picture (or touch the picture on your tablet), and you'll see this very corner X.

The cosine of which angle is 0.5?

x = π /3

cos 60°= cos( π /3) = 0,5

Some people will chuckle skeptically, yes... Like, was it worth making a circle when everything is already clear... You can, of course, chuckle...) But the fact is that this is an erroneous answer. Or rather, insufficient. Circle connoisseurs understand that there are a whole bunch of other angles here that also give a cosine of 0.5.

If you turn the moving side OA full turn, point A will return to its original position. With the same cosine equal to 0.5. Those. the angle will change by 360° or 2π radians, and cosine - no. The new angle 60° + 360° = 420° will also be a solution to our equation, because

An infinite number of such complete revolutions can be made... And all these new angles will be solutions to our trigonometric equation. And they all need to be written down somehow in response. All. Otherwise, the decision does not count, yes...)

Mathematics can do this simply and elegantly. Write down in one short answer infinite set decisions. Here's what it looks like for our equation:

x = π /3 + 2π n, n ∈ Z

I'll decipher it. Still write meaningfully It’s more pleasant than stupidly drawing some mysterious letters, right?)

π /3 - this is the same corner that we saw on the circle and determined according to the cosine table.

2π is one complete revolution in radians.

n - this is the number of complete ones, i.e. whole rpm It is clear that n can be equal to 0, ±1, ±2, ±3.... and so on. As stated short note:

n ∈ Z

n belongs ( ∈ ) set of integers ( Z ). By the way, instead of the letter n letters may well be used k, m, t etc.

This notation means you can take any integer n . At least -3, at least 0, at least +55. Whatever you want. If you substitute this number into the answer, you will get a specific angle, which will definitely be the solution to our harsh equation.)

Or, in other words, x = π /3 is the only root of infinite number. To get all the other roots, it is enough to add any number of full revolutions to π /3 ( n ) in radians. Those. 2πn radian.

All? No. I deliberately prolong the pleasure. To remember better.) We received only part of the answers to our equation. I will write this first part of the solution like this:

x 1 = π /3 + 2π n, n ∈ Z

x 1 - not just one root, but a whole series of roots, written down in a short form.

But there are also angles that also give a cosine of 0.5!

Let's return to our picture from which we wrote down the answer. Here she is:

Hover your mouse over the image and we see another angle that also gives a cosine of 0.5. What do you think it is equal to? The triangles are the same... Yes! It is equal to the angle X , only delayed in the negative direction. This is the corner -X. But we have already calculated x. π /3 or 60°. Therefore, we can safely write:

x 2 = - π /3

Well, of course, we add all the angles that are obtained through full revolutions:

x 2 = - π /3 + 2π n, n ∈ Z

That's all now.) On the trigonometric circle we saw(who understands, of course)) All angles that give a cosine of 0.5. And we wrote down these angles in a short mathematical form. The answer resulted in two infinite series of roots:

x 1 = π /3 + 2π n, n ∈ Z

x 2 = - π /3 + 2π n, n ∈ Z

This is the correct answer.

Hope, general principle for solving trigonometric equations using a circle is clear. We mark on the circle the cosine (sine, tangent, cotangent) from given equation, draw the corresponding angles and write down the answer. Of course, we need to figure out what corners we are saw on the circle. Sometimes it's not so obvious. Well, I said that logic is required here.)

For example, let's look at another trigonometric equation:

Please take into account that the number 0.5 is not the only possible number in equations!) It’s just more convenient for me to write it than roots and fractions.

We work according to the general principle. We draw a circle, mark (on the sine axis, of course!) 0.5. We draw all the angles corresponding to this sine at once. We get this picture:

Let's deal with the angle first X in the first quarter. We recall the table of sines and determine the value of this angle. It's a simple matter:

x = π /6

We remember about full revolutions and, with clear conscience, we write down the first series of answers:

x 1 = π /6 + 2π n, n ∈ Z

Half the job is done. But now we need to determine second corner... It's trickier than using cosines, yes... But logic will save us! How to determine the second angle through x? Yes Easy! The triangles in the picture are the same, and the red corner X equal to angle X . Only it is counted from the angle π in the negative direction. That’s why it’s red.) And for the answer we need an angle, measured correctly, from the positive semi-axis OX, i.e. from an angle of 0 degrees.

We hover the cursor over the drawing and see everything. I removed the first corner so as not to complicate the picture. The angle we are interested in (drawn in green) will be equal to:

π - x

X we know this π /6 . Therefore, the second angle will be:

π - π /6 = 5π /6

Again we remember about adding full revolutions and write down the second series of answers:

x 2 = 5π /6 + 2π n, n ∈ Z

That's all. A complete answer consists of two series of roots:

x 1 = π /6 + 2π n, n ∈ Z

x 2 = 5π /6 + 2π n, n ∈ Z

Tangent and cotangent equations can be easily solved using the same general principle for solving trigonometric equations. If, of course, you know how to draw tangent and cotangent on a trigonometric circle.

In the examples above, I used the table value of sine and cosine: 0.5. Those. one of those meanings that the student knows must. Now let's expand our capabilities to all other values. Decide, so decide!)

So, let's say we need to solve this trigonometric equation:

There is no such cosine value in the short tables. We coldly ignore this terrible fact. Draw a circle, mark 2/3 on the cosine axis and draw the corresponding angles. We get this picture.

Let's look, first, at the angle in the first quarter. If only we knew what x is equal to, we would immediately write down the answer! We don’t know... Failure!? Calm! Mathematics does not leave its own people in trouble! She came up with arc cosines for this case. Do not know? In vain. Find out, It's a lot easier than you think. On this link there is not a single tricky spell about “reverse trigonometric functions“No... This is superfluous in this topic.

If you are in the know, just say to yourself: “X is an angle whose cosine is equal to 2/3.” And immediately, purely by the definition of arc cosine, we can write:

We remember about the additional revolutions and calmly write down the first series of roots of our trigonometric equation:

x 1 = arccos 2/3 + 2π n, n ∈ Z

The second series of roots for the second angle is almost automatically written down. Everything is the same, only X (arccos 2/3) will be with a minus:

x 2 = - arccos 2/3 + 2π n, n ∈ Z

And that's it! This is the correct answer. Even easier than with table values. There is no need to remember anything.) By the way, the most attentive will notice that this picture shows the solution through the arc cosine in essence, no different from the picture for the equation cosx = 0.5.

Exactly! General principle That's why it's common! I deliberately drew two almost identical pictures. The circle shows us the angle X by its cosine. Whether it is a tabular cosine or not is unknown to everyone. What kind of angle this is, π /3, or what arc cosine is - that’s up to us to decide.

Same song with sine. For example:

Draw a circle again, mark the sine equal to 1/3, draw the angles. This is the picture we get:

And again the picture is almost the same as for the equation sinx = 0.5. Again we start from the corner in the first quarter. What is X equal to if its sine is 1/3? No problem!

Now the first pack of roots is ready:

x 1 = arcsin 1/3 + 2π n, n ∈ Z

Let's deal with the second angle. In the example with a table value of 0.5, it was equal to:

π - x

It will be exactly the same here too! Only x is different, arcsin 1/3. So what!? You can safely write down the second pack of roots:

x 2 = π - arcsin 1/3 + 2π n, n ∈ Z

This is a completely correct answer. Although it doesn't look very familiar. But it’s clear, I hope.)

This is how trigonometric equations are solved using a circle. This path is clear and understandable. It is he who saves in trigonometric equations with selection of roots on a given interval, in trigonometric inequalities- those are generally resolved almost always in a circle. In short, in any tasks that are a little more difficult than standard ones.

Let's apply knowledge in practice?)

Solve trigonometric equations:

First, simpler, straight from this lesson.

Now it's more complicated.

Hint: here you will have to think about the circle. Personally.)

And now they are outwardly simple... They are also called special cases.

sinx = 0

sinx = 1

cosx = 0

cosx = -1

Hint: here you need to figure out in a circle where there are two series of answers and where there is one... And how to write one instead of two series of answers. Yes, so that not a single root from an infinite number is lost!)

Well, very simple):

sinx = 0,3

cosx = π

tgx = 1,2

ctgx = 3,7

Hint: here you need to know what arcsine and arccosine are? What is arctangent, arccotangent? The simplest definitions. But you don’t need to remember any table values!)

The answers are, of course, a mess):

x 1= arcsin0,3 + 2π n, n ∈ Z
x 2= π - arcsin0.3 + 2

Not everything works out? Happens. Read the lesson again. Only thoughtfully(there is such obsolete word...) And follow the links. The main links are about the circle. Without it, trigonometry is like crossing the road blindfolded. Sometimes it works.)

If you like this site...

By the way, I have a couple more interesting sites for you.)

You can practice solving examples and find out your level. Testing with instant verification. Let's learn - with interest!)

You can get acquainted with functions and derivatives.

Solving simple trigonometric equations.

Solving trigonometric equations of any level of complexity ultimately comes down to solving the simplest trigonometric equations. And in this the trigonometric circle again turns out to be the best assistant.

Let's recall the definitions of cosine and sine.

The cosine of an angle is the abscissa (that is, the coordinate along the axis) of a point on the unit circle corresponding to a rotation through a given angle.

The sine of an angle is the ordinate (that is, the coordinate along the axis) of a point on the unit circle corresponding to a rotation through a given angle.

The positive direction of movement on the trigonometric circle is counterclockwise. A rotation of 0 degrees (or 0 radians) corresponds to a point with coordinates (1;0)

We use these definitions to solve simple trigonometric equations.

1. Solve the equation

This equation is satisfied by all values ​​of the rotation angle that correspond to points on the circle whose ordinate is equal to .

Let's mark a point with ordinate on the ordinate axis:

Draw a horizontal line parallel to the x-axis until it intersects with the circle. We get two points lying on the circle and having an ordinate. These points correspond to rotation angles in and radians:

If we, leaving the point corresponding to the angle of rotation per radian, go around a full circle, then we will arrive at a point corresponding to the angle of rotation per radian and having the same ordinate. That is, this rotation angle also satisfies our equation. We can make as many “idle” revolutions as we like, returning to the same point, and all these angle values ​​will satisfy our equation. The number of “idle” revolutions will be denoted by the letter (or). Since we can make these revolutions in both positive and negative directions, (or) can take on any integer values.

That is, the first series of solutions to the original equation has the form:

, , - set of integers (1)

Similarly, the second series of solutions has the form:

, Where , . (2)

As you might have guessed, this series of solutions is based on the point on the circle corresponding to the angle of rotation by .

These two series of solutions can be combined into one entry:

If we take (that is, even) in this entry, then we will get the first series of solutions.

If we take (that is, odd) in this entry, then we get the second series of solutions.

2. Now let's solve the equation

Since this is the abscissa of a point on the unit circle obtained by rotating through an angle, we mark the point with the abscissa on the axis:

Draw a vertical line parallel to the axis until it intersects with the circle. We will get two points lying on the circle and having an abscissa. These points correspond to rotation angles in and radians. Recall that when moving clockwise we get a negative rotation angle:

Let us write down two series of solutions:

,

,

(We get to the desired point by going from the main full circle, that is.

Let's combine these two series into one entry:

3. Solve the equation

The tangent line passes through the point with coordinates (1,0) of the unit circle parallel to the OY axis

Let's mark a point on it with an ordinate equal to 1 (we are looking for the tangent of which angles is equal to 1):

Let's connect this point to the origin of coordinates with a straight line and mark the points of intersection of the line with the unit circle. The intersection points of the straight line and the circle correspond to the angles of rotation on and :

Since the points corresponding to the rotation angles that satisfy our equation lie at a distance of radians from each other, we can write the solution this way:

4. Solve the equation

The line of cotangents passes through the point with the coordinates of the unit circle parallel to the axis.

Let's mark a point with abscissa -1 on the line of cotangents:

Let's connect this point to the origin of the straight line and continue it until it intersects with the circle. This straight line will intersect the circle at points corresponding to the angles of rotation in and radians:

Since these points are separated from each other by a distance equal to , then common decision We can write this equation like this:

In the given examples illustrating the solution of the simplest trigonometric equations, tabular values ​​of trigonometric functions were used.

However, if the right side of the equation contains a non-tabular value, then we substitute the value into the general solution of the equation:

SPECIAL SOLUTIONS:

Let us mark the points on the circle whose ordinate is 0:

Let us mark a single point on the circle whose ordinate is 1:

Let us mark a single point on the circle whose ordinate is equal to -1:

Since it is customary to indicate values ​​closest to zero, we write the solution as follows:

Let us mark the points on the circle whose abscissa is equal to 0:

5.
Let us mark a single point on the circle whose abscissa is equal to 1:

Let us mark a single point on the circle whose abscissa is equal to -1:

And slightly more complex examples:

1.

The sine is equal to one if the argument is equal to

The argument of our sine is equal, so we get:

Let's divide both sides of the equality by 3:

Answer:

2.

Cosine is zero if the argument of cosine is

The argument of our cosine is equal to , so we get:

Let's express , to do this we first move to the right with the opposite sign:

Let's simplify the right side:

Divide both sides by -2:

Note that the sign in front of the term does not change, since k can take any integer value.

Answer:

And finally, watch the video lesson “Selecting roots in a trigonometric equation using a trigonometric circle”

This concludes our conversation about solving simple trigonometric equations. Next time we will talk about how to decide.

I once witnessed a conversation between two applicants:

– When should you add 2πn, and when should you add πn? I just can't remember!

– And I have the same problem.

I just wanted to tell them: “You don’t need to memorize, but understand!”

This article is addressed primarily to high school students and, I hope, will help them solve the simplest trigonometric equations with “understanding”:

Number circle

Along with the concept of a number line, there is also the concept of a number circle. As we know, in a rectangular coordinate system, a circle with a center at the point (0;0) and radius 1 is called a unit circle. Let’s imagine the number line as a thin thread and wind it around this circle: we will attach the origin (point 0) to the “right” point of the unit circle, we will wrap the positive semi-axis counterclockwise, and the negative semi-axis in the direction (Fig. 1). Such a unit circle is called a numerical circle.

Properties of the number circle

• Each real number lies on one point on the number circle.
• There are infinitely many real numbers at every point on the number circle. Since the length of the unit circle is 2π, the difference between any two numbers at one point on the circle is equal to one of the numbers ±2π; ±4π ; ±6π ; ...

Let's conclude: knowing one of the numbers of point A, we can find all the numbers of point A.

Let's draw the diameter of the AC (Fig. 2). Since x_0 is one of the numbers of point A, then the numbers x_0±π ; x_0±3π; x_0±5π; ... and only they will be the numbers of point C. Let's choose one of these numbers, say, x_0+π, and use it to write down all the numbers of point C: x_C=x_0+π+2πk ,k∈Z. Note that the numbers at points A and C can be combined into one formula: x_(A ; C)=x_0+πk ,k∈Z (for k = 0; ±2; ±4; ... we obtain the numbers of point A, and for k = ±1; ±3; ±5; … – numbers of point C).

Let's conclude: knowing one of the numbers at one of the points A or C of the diameter AC, we can find all the numbers at these points.

• Two opposite numbers are located on points of the circle that are symmetrical with respect to the abscissa axis.

Let's draw a vertical chord AB (Fig. 2). Since points A and B are symmetrical about the Ox axis, the number -x_0 is located at point B and, therefore, all numbers of point B are given by the formula: x_B=-x_0+2πk ,k∈Z. We write the numbers at points A and B using one formula: x_(A ; B)=±x_0+2πk ,k∈Z. Let us conclude: knowing one of the numbers at one of the points A or B of the vertical chord AB, we can find all the numbers at these points. Let's consider the horizontal chord AD and find the numbers of point D (Fig. 2). Since BD is a diameter and the number -x_0 belongs to point B, then -x_0 + π is one of the numbers of point D and, therefore, all the numbers of this point are given by the formula x_D=-x_0+π+2πk ,k∈Z. The numbers at points A and D can be written using one formula: x_(A ; D)=(-1)^k∙x_0+πk ,k∈Z . (for k= 0; ±2; ±4; … we get the numbers of point A, and for k = ±1; ±3; ±5; … – the numbers of point D).

Let's conclude: knowing one of the numbers at one of the points A or D of the horizontal chord AD, we can find all the numbers at these points.

Sixteen main points of the number circle

In practice, solving most of the simplest trigonometric equations involves sixteen points on a circle (Fig. 3). What are these dots? Red, blue and green dots divide the circle into 12 equal parts. Since the length of the semicircle is π, then the length of the arc A1A2 is π/2, the length of the arc A1B1 is π/6, and the length of the arc A1C1 is π/3.

Now we can indicate one number at a time:

π/3 on C1 and

The vertices of the orange square are the midpoints of the arcs of each quarter, therefore, the length of the arc A1D1 is equal to π/4 and, therefore, π/4 is one of the numbers of point D1. Using the properties of the number circle, we can use formulas to write down all the numbers on all marked points of our circle. The coordinates of these points are also marked in the figure (we will omit the description of their acquisition).

Having learned the above, we now have sufficient preparation to solve special cases (for nine values ​​of the number a) simplest equations.

Solve equations

1)sinx=1⁄(2).

– What is required of us?

– Find all those numbers x whose sine is 1/2.

Let's remember the definition of sine: sinx – ordinate of the point on the number circle on which the number x is located. We have two points on the circle whose ordinate is equal to 1/2. These are the ends of the horizontal chord B1B2. This means that the requirement “solve the equation sinx=1⁄2” is equivalent to the requirement “find all the numbers at point B1 and all the numbers at point B2.”

2)sinx=-√3⁄2 .

We need to find all the numbers at points C4 and C3.

3) sinx=1. On the circle we have only one point with ordinate 1 - point A2 and, therefore, we need to find only all the numbers of this point.

Answer: x=π/2+2πk, k∈Z.

4)sinx=-1 .

Only point A_4 has an ordinate of -1. All the numbers of this point will be the horses of the equation.

Answer: x=-π/2+2πk, k∈Z.

5) sinx=0 .

On the circle we have two points with ordinate 0 - points A1 and A3. You can indicate the numbers at each of the points separately, but given that these points are diametrically opposite, it is better to combine them into one formula: x=πk,k∈Z.

Answer: x=πk ,k∈Z .

6)cosx=√2⁄2 .

Let's remember the definition of cosine: cosx is the abscissa of the point on the number circle on which the number x is located. On the circle we have two points with the abscissa √2⁄2 - the ends of the horizontal chord D1D4. We need to find all the numbers on these points. Let's write them down, combining them into one formula.

Answer: x=±π/4+2πk, k∈Z.

7) cosx=-1⁄2 .

We need to find the numbers at points C_2 and C_3.

Answer: x=±2π/3+2πk , k∈Z .

10) cosx=0 .

Only points A2 and A4 have an abscissa of 0, which means that all the numbers at each of these points will be solutions to the equation.
.

The solutions to the equation of the system are the numbers at points B_3 and B_4. To the cosx inequality<0 удовлетворяют только числа b_3
Answer: x=-5π/6+2πk, k∈Z.

Note that for any admissible value of x, the second factor is positive and, therefore, the equation is equivalent to the system

The solutions to the system equation are the number of points D_2 and D_3. The numbers of point D_2 do not satisfy the inequality sinx≤0.5, but the numbers of point D_3 do.

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The simplest trigonometric equations are solved, as a rule, using formulas. Let me remind you that the simplest trigonometric equations are:

sinx = a

cosx = a

tgx = a

ctgx = a

x is the angle to be found,
a is any number.

And here are the formulas with which you can immediately write down the solutions to these simplest equations.

For sine:

For cosine:

x = ± arccos a + 2π n, n ∈ Z

For tangent:

x = arctan a + π n, n ∈ Z

For cotangent:

x = arcctg a + π n, n ∈ Z

Actually, this is the theoretical part of solving the simplest trigonometric equations. Moreover, everything!) Nothing at all. However, the number of errors on this topic is simply off the charts. Especially if the example deviates slightly from the template. Why?

Yes, because a lot of people write down these letters, without understanding their meaning at all! He writes down with caution, lest something happen...) This needs to be sorted out. Trigonometry for people, or people for trigonometry, after all!?)

Let's figure it out?

One angle will be equal to arccos a, second: -arccos a.

And it will always work out this way. For any A.

If you don’t believe me, hover your mouse over the picture, or touch the picture on your tablet.) I changed the number A to something negative. Anyway, we got one corner arccos a, second: -arccos a.

Therefore, the answer can always be written as two series of roots:

x 1 = arccos a + 2π n, n ∈ Z

x 2 = - arccos a + 2π n, n ∈ Z

Let's combine these two series into one:

x= ± arccos a + 2π n, n ∈ Z

And that's all. We have obtained a general formula for solving the simplest trigonometric equation with cosine.

If you understand that this is not some kind of superscientific wisdom, but just a shortened version of two series of answers, You will also be able to handle tasks “C”. With inequalities, with selecting roots from a given interval... There the answer with a plus/minus does not work. But if you treat the answer in a businesslike manner and break it down into two separate answers, everything will be resolved.) Actually, that’s why we’re looking into it. What, how and where.

In the simplest trigonometric equation

sinx = a

we also get two series of roots. Always. And these two series can also be recorded in one line. Only this line will be trickier:

x = (-1) n arcsin a + π n, n ∈ Z

But the essence remains the same. Mathematicians simply designed a formula to make one instead of two entries for series of roots. That's all!

Let's check the mathematicians? And you never know...)

In the previous lesson, the solution (without any formulas) of a trigonometric equation with sine was discussed in detail:

The answer resulted in two series of roots:

x 1 = π /6 + 2π n, n ∈ Z

x 2 = 5π /6 + 2π n, n ∈ Z

If we solve the same equation using the formula, we get the answer:

x = (-1) n arcsin 0.5 + π n, n ∈ Z

Actually, this is an unfinished answer.) The student must know that arcsin 0.5 = π /6. The complete answer would be:

x = (-1)n π /6+ π n, n ∈ Z

This raises an interesting question. Reply via x 1; x 2 (this is the correct answer!) and through lonely X (and this is the correct answer!) - are they the same thing or not? We'll find out now.)

We substitute in the answer with x 1 values n =0; 1; 2; etc., we count, we get a series of roots:

x 1 = π/6; 13π/6; 25π/6 and so on.

With the same substitution in response with x 2 , we get:

x 2 = 5π/6; 17π/6; 29π/6 and so on.

Now let's substitute the values n (0; 1; 2; 3; 4...) into the general formula for single X . That is, we raise minus one to the zero power, then to the first, second, etc. Well, of course, we substitute 0 into the second term; 1; 2 3; 4, etc. And we count. We get the series:

x = π/6; 5π/6; 13π/6; 17π/6; 25π/6 and so on.

That's all you can see.) The general formula gives us exactly the same results as are the two answers separately. Just everything at once, in order. The mathematicians were not fooled.)

Formulas for solving trigonometric equations with tangent and cotangent can also be checked. But we won’t.) They are already simple.

I wrote out all this substitution and checking specifically. Here it is important to understand one simple thing: there are formulas for solving elementary trigonometric equations, just a short summary of the answers. For this brevity, we had to insert plus/minus into the cosine solution and (-1) n into the sine solution.

These inserts do not interfere in any way in tasks where you just need to write down the answer to an elementary equation. But if you need to solve an inequality, or then you need to do something with the answer: select roots on an interval, check for ODZ, etc., these insertions can easily unsettle a person.

So what should I do? Yes, either write the answer in two series, or solve the equation/inequality using the trigonometric circle. Then these insertions disappear and life becomes easier.)

We can summarize.

To solve the simplest trigonometric equations, there are ready-made answer formulas. Four pieces. They are good for instantly writing down the solution to an equation. For example, you need to solve the equations:

sinx = 0.3

Easily: x = (-1) n arcsin 0.3 + π n, n ∈ Z

cosx = 0.2

No problem: x = ± arccos 0.2 + 2π n, n ∈ Z

tgx = 1.2

Easily: x = arctan 1,2 + π n, n ∈ Z

ctgx = 3.7

One left: x= arcctg3,7 + π n, n ∈ Z

cos x = 1.8

If you, shining with knowledge, instantly write the answer:

x= ± arccos 1.8 + 2π n, n ∈ Z

then you are already shining, this... that... from a puddle.) Correct answer: there are no solutions. Don't understand why? Read what arc cosine is. In addition, if on the right side of the original equation there are tabular values ​​of sine, cosine, tangent, cotangent, - 1; 0; √3; 1/2; √3/2 and so on. - the answer through the arches will be unfinished. Arches must be converted to radians.

And if you come across inequality, like

then the answer is:

x πn, n ∈ Z

there is rare nonsense, yes...) Here you need to solve using the trigonometric circle. What we will do in the corresponding topic.

For those who heroically read to these lines. I simply cannot help but appreciate your titanic efforts. Bonus for you.)

Bonus:

When writing down formulas in an alarming combat situation, even seasoned nerds often get confused about where πn, And where 2π n. Here's a simple trick for you. In everyone formulas worth πn. Except for the only formula with arc cosine. It stands there 2πn. Two peen. Keyword - two. In this same formula there are two sign at the beginning. Plus and minus. Here and there - two.

So if you wrote two sign before the arc cosine, it’s easier to remember what will happen at the end two peen. And it also happens the other way around. The person will miss the sign ± , gets to the end, writes correctly two Pien, and he’ll come to his senses. There's something ahead two sign! The person will return to the beginning and correct the mistake! Like this.)

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