Examples of solutions to complex trigonometric equations. Basic methods for solving trigonometric equations

Methods for solving trigonometric equations

Introduction 2

Methods for solving trigonometric equations 5

Algebraic 5

Solving equations using the condition of equality of trigonometric functions of the same name 7

Factorization 8

Reduction to homogeneous equation 10

Introduction of auxiliary angle 11

Convert product to sum 14

Universal substitution 14

Conclusion 17

Introduction

Until the tenth grade, the order of actions of many exercises leading to the goal is, as a rule, clearly defined. For example, linear and quadratic equations and inequalities, fractional equations and equations reducible to quadratic ones, etc. Without examining in detail the principle of solving each of the mentioned examples, we note the general things that are necessary for their successful solution.

In most cases, you need to establish what type of task the task is, remember the sequence of actions leading to the goal, and perform these actions. Obviously, the success or failure of a student in mastering techniques for solving equations depends mainly on how well he is able to correctly determine the type of equation and remember the sequence of all stages of its solution. Of course, it is assumed that the student has the skills to perform identical transformations and calculations.

A completely different situation arises when a schoolchild encounters trigonometric equations. Moreover, it is not difficult to establish the fact that the equation is trigonometric. Difficulties arise when finding a course of action that would lead to a positive result. And here the student faces two problems. It is difficult to determine the type by the appearance of the equation. And without knowing the type, it is almost impossible to select the desired formula from the several dozen available.

To help students find their way through the complex maze of trigonometric equations, they are first introduced to equations that are reduced to quadratic equations when a new variable is introduced. Then they solve homogeneous equations and those reducible to them. Everything ends, as a rule, with equations, to solve which it is necessary to factor the left-hand side, then equating each of the factors to zero.

Realizing that the dozen and a half equations discussed in lessons are clearly not enough to set the student on an independent voyage through the trigonometric “sea,” the teacher adds a few more recommendations of his own.

To solve a trigonometric equation, you need to try:

Bring all functions included in the equation to “the same angles”;

Reduce the equation to “identical functions”;

Factor the left side of the equation, etc.

But despite knowing the basic types of trigonometric equations and several principles for finding their solutions, many students still find themselves stumped by every equation that is slightly different from those solved before. It remains unclear what one should strive for when having this or that equation, why in one case it is necessary to use double angle formulas, in another - half angle, and in a third - addition formulas, etc.

Definition 1. A trigonometric equation is an equation in which the unknown is contained under the sign of trigonometric functions.

Definition 2. A trigonometric equation is said to have equal angles if all trigonometric functions included in it have equal arguments. A trigonometric equation is said to have identical functions if it contains only one of the trigonometric functions.

Definition 3. The power of a monomial containing trigonometric functions is the sum of the exponents of the powers of the trigonometric functions included in it.

Definition 4. An equation is called homogeneous if all monomials included in it have the same degree. This degree is called the order of the equation.

Definition 5. Trigonometric equation containing only functions sin And cos, is called homogeneous if all monomials with respect to trigonometric functions have the same degree, and the trigonometric functions themselves have equal angles and the number of monomials is 1 greater than the order of the equation.

Methods for solving trigonometric equations.

Solving trigonometric equations consists of two stages: transforming the equation to obtain its simplest form and solving the resulting simplest trigonometric equation. There are seven basic methods for solving trigonometric equations.

I. Algebraic method. This method is well known from algebra. (Method of variable replacement and substitution).

Solve equations.

1)

Let us introduce the notation x=2 sin3 t, we get

Solving this equation, we get:
or

those. can be written down

When recording the resulting solution due to the presence of signs degree
there is no point in writing it down.

Answer:

Let's denote

We get a quadratic equation
. Its roots are numbers
And
. Therefore, this equation reduces to the simplest trigonometric equations
And
. Solving them, we find that
or
.

Answer:
;
.

Let's denote

does not satisfy the condition

Means

Answer:

Let's transform the left side of the equation:

Thus, this initial equation can be written as:

, i.e.

Having designated
, we get
Solving this quadratic equation we have:

does not satisfy the condition

We write down the solution to the original equation:

Answer:

Substitution
reduces this equation to a quadratic equation
. Its roots are numbers
And
. Because
, then the given equation has no roots.

Answer: no roots.

II. Solving equations using the condition of equality of trigonometric functions of the same name.

A)
, If

b)
, If

V)
, If

Using these conditions, consider solving the following equations:

6)

Using what was said in part a) we find that the equation has a solution if and only if
.

Solving this equation, we find
.

We have two groups of solutions:

.

7) Solve the equation:
.

Using the condition of item b) we deduce that
.

Solving these quadratic equations, we get:

.

8) Solve the equation
.

From this equation we deduce that . Solving this quadratic equation, we find that

.

III. Factorization.

We consider this method with examples.

9) Solve the equation
.

Solution. Let's move all the terms of the equation to the left: .

Let's transform and factorize the expression on the left side of the equation:
.

.

.

1)
2)

Because
And
do not accept the value zero

at the same time, then we divide both parts

equations for
,

Answer:

10) Solve the equation:

Solution.

or

Answer:

11) Solve the equation

Solution:

1)
2)
3)

,

Answer:

IV. Reduction to a homogeneous equation.

To solve a homogeneous equation you need:

Move all its members to the left side;

Place all common factors out of brackets;

Equate all factors and brackets to zero;

Brackets equal to zero give a homogeneous equation of lesser degree, which should be divided by
(or
) in the senior degree;

Solve the resulting algebraic equation for
.

Let's look at examples:

12) Solve the equation:

Solution.

Let's divide both sides of the equation by
,

Introducing designations
, name

roots of this equation:

hence 1)
2)

Answer:

13) Solve the equation:

Solution. Using the double angle formulas and the basic trigonometric identity, we reduce this equation to a half argument:

After reducing similar terms we have:

Dividing the homogeneous last equation by
, we get

I will indicate
, we get a quadratic equation
, whose roots are numbers

Thus

Expression
goes to zero at
, i.e. at
,
.

The solution to the equation we obtained does not include these numbers.

Answer:
, .

V. Introduction of an auxiliary angle.

Consider an equation of the form

Where a, b, c- coefficients, x- unknown.

Let's divide both sides of this equation by

Now the coefficients of the equation have the properties of sine and cosine, namely: the modulus of each of them does not exceed one, and the sum of their squares is equal to 1.

Then we can designate them accordingly
(Here - auxiliary angle) and our equation takes the form: .

Then

And his decision

Note that the introduced notations are mutually interchangeable.

14) Solve the equation:

Solution. Here
, so we divide both sides of the equation by

Answer:

15) Solve the equation

Solution. Because
, then this equation is equivalent to the equation

Because
, then there is an angle such that
,
(those.
).

We have

Because
, then we finally get:

.

Note that equations of the form have a solution if and only if

16) Solve the equation:

To solve this equation, we group trigonometric functions with the same arguments

Divide both sides of the equation by two

Let's transform the sum of trigonometric functions into a product:

Answer:

VI. Converting a product to a sum.

The corresponding formulas are used here.

17) Solve the equation:

Solution. Let's transform the left side into a sum:

VII.Universal substitution.

,

these formulas are true for everyone

Substitution
called universal.

18) Solve the equation:

Solution: Replace and
to their expression through
and denote
.

We get a rational equation
, which converts to square
.

The roots of this equation are the numbers
.

Therefore, the problem was reduced to solving two equations
.

We find that
.

View value
does not satisfy the original equation, which is verified by checking - substituting the given value t into the original equation.

Answer:
.

Comment. Equation 18 could have been solved in another way.

Let's divide both sides of this equation by 5 (i.e. by
):
.

Because
, then there is such a number
, What
And
. Therefore the equation takes the form:
or
. From here we find that
Where
.

19) Solve the equation
.

Solution. Since the functions
And
have the greatest value equal to 1, then their sum is equal to 2 if
And
, simultaneously, that is
.

Answer:
.

When solving this equation, the boundedness of the functions and was used.

Conclusion.

When working on the topic “Solving trigonometric equations,” it is useful for each teacher to follow the following recommendations:

Systematize methods for solving trigonometric equations.

Choose for yourself the steps to perform an analysis of the equation and signs of the advisability of using a particular solution method.

Think over ways to self-monitor your activities in implementing the method.

Learn to compose “your own” equations for each of the methods being studied.

Appendix No. 1

Solve homogeneous or reducible to homogeneous equations.

1.	Rep.
	Rep.
	Rep.
5.	Rep.
	Rep.
7.	Rep.
	Rep.

When solving many mathematical problems, especially those that occur before grade 10, the order of actions performed that will lead to the goal is clearly defined. Such problems include, for example, linear and quadratic equations, linear and quadratic inequalities, fractional equations and equations that reduce to quadratic ones. The principle of successfully solving each of the mentioned problems is as follows: you need to establish what type of problem you are solving, remember the necessary sequence of actions that will lead to the desired result, i.e. answer and follow these steps.

It is obvious that success or failure in solving a particular problem depends mainly on how correctly the type of equation being solved is determined, how correctly the sequence of all stages of its solution is reproduced. Of course, in this case it is necessary to have the skills to perform identical transformations and calculations.

The situation is different with trigonometric equations. It is not at all difficult to establish the fact that the equation is trigonometric. Difficulties arise when determining the sequence of actions that would lead to the correct answer.

It is sometimes difficult to determine its type based on the appearance of an equation. And without knowing the type of equation, it is almost impossible to choose the right one from several dozen trigonometric formulas.

To solve a trigonometric equation, you need to try:

1. bring all functions included in the equation to “the same angles”;
2. bring the equation to “identical functions”;
3. factor the left side of the equation, etc.

Let's consider basic methods for solving trigonometric equations.

I. Reduction to the simplest trigonometric equations

Solution diagram

Step 1. Express a trigonometric function in terms of known components.

Step 2. Find the function argument using the formulas:

cos x = a; x = ±arccos a + 2πn, n ЄZ.

sin x = a; x = (-1) n arcsin a + πn, n Є Z.

tan x = a; x = arctan a + πn, n Є Z.

ctg x = a; x = arcctg a + πn, n Є Z.

Step 3. Find the unknown variable.

Example.

2 cos(3x – π/4) = -√2.

Solution.

1) cos(3x – π/4) = -√2/2.

2) 3x – π/4 = ±(π – π/4) + 2πn, n Є Z;

3x – π/4 = ±3π/4 + 2πn, n Є Z.

3) 3x = ±3π/4 + π/4 + 2πn, n Є Z;

x = ±3π/12 + π/12 + 2πn/3, n Є Z;

x = ±π/4 + π/12 + 2πn/3, n Є Z.

Answer: ±π/4 + π/12 + 2πn/3, n Є Z.

II. Variable replacement

Solution diagram

Step 1. Reduce the equation to algebraic form with respect to one of the trigonometric functions.

Step 2. Denote the resulting function by the variable t (if necessary, introduce restrictions on t).

Step 3. Write down and solve the resulting algebraic equation.

Step 4. Make a reverse replacement.

Step 5. Solve the simplest trigonometric equation.

Example.

2cos 2 (x/2) – 5sin (x/2) – 5 = 0.

Solution.

1) 2(1 – sin 2 (x/2)) – 5sin (x/2) – 5 = 0;

2sin 2 (x/2) + 5sin (x/2) + 3 = 0.

2) Let sin (x/2) = t, where |t| ≤ 1.

3) 2t 2 + 5t + 3 = 0;

t = 1 or e = -3/2, does not satisfy the condition |t| ≤ 1.

4) sin(x/2) = 1.

5) x/2 = π/2 + 2πn, n Є Z;

x = π + 4πn, n Є Z.

Answer: x = π + 4πn, n Є Z.

III. Equation order reduction method

Solution diagram

Step 1. Replace this equation with a linear one, using the formula for reducing the degree:

sin 2 x = 1/2 · (1 – cos 2x);

cos 2 x = 1/2 · (1 + cos 2x);

tg 2 x = (1 – cos 2x) / (1 + cos 2x).

Step 2. Solve the resulting equation using methods I and II.

Example.

cos 2x + cos 2 x = 5/4.

Solution.

1) cos 2x + 1/2 · (1 + cos 2x) = 5/4.

2) cos 2x + 1/2 + 1/2 · cos 2x = 5/4;

3/2 cos 2x = 3/4;

2x = ±π/3 + 2πn, n Є Z;

x = ±π/6 + πn, n Є Z.

Answer: x = ±π/6 + πn, n Є Z.

IV. Homogeneous equations

Solution diagram

Step 1. Reduce this equation to the form

a) a sin x + b cos x = 0 (homogeneous equation of the first degree)

or to the view

b) a sin 2 x + b sin x · cos x + c cos 2 x = 0 (homogeneous equation of the second degree).

Step 2. Divide both sides of the equation by

a) cos x ≠ 0;

b) cos 2 x ≠ 0;

and get the equation for tan x:

a) a tan x + b = 0;

b) a tan 2 x + b arctan x + c = 0.

Step 3. Solve the equation using known methods.

Example.

5sin 2 x + 3sin x cos x – 4 = 0.

Solution.

1) 5sin 2 x + 3sin x · cos x – 4(sin 2 x + cos 2 x) = 0;

5sin 2 x + 3sin x · cos x – 4sin² x – 4cos 2 x = 0;

sin 2 x + 3sin x · cos x – 4cos 2 x = 0/cos 2 x ≠ 0.

2) tg 2 x + 3tg x – 4 = 0.

3) Let tg x = t, then

t 2 + 3t – 4 = 0;

t = 1 or t = -4, which means

tg x = 1 or tg x = -4.

From the first equation x = π/4 + πn, n Є Z; from the second equation x = -arctg 4 + πk, k Є Z.

Answer: x = π/4 + πn, n Є Z; x = -arctg 4 + πk, k Є Z.

V. Method of transforming an equation using trigonometric formulas

Solution diagram

Step 1. Using all possible trigonometric formulas, reduce this equation to an equation solved by methods I, II, III, IV.

Step 2. Solve the resulting equation using known methods.

Example.

sin x + sin 2x + sin 3x = 0.

Solution.

1) (sin x + sin 3x) + sin 2x = 0;

2sin 2x cos x + sin 2x = 0.

2) sin 2x (2cos x + 1) = 0;

sin 2x = 0 or 2cos x + 1 = 0;

From the first equation 2x = π/2 + πn, n Є Z; from the second equation cos x = -1/2.

We have x = π/4 + πn/2, n Є Z; from the second equation x = ±(π – π/3) + 2πk, k Є Z.

As a result, x = π/4 + πn/2, n Є Z; x = ±2π/3 + 2πk, k Є Z.

Answer: x = π/4 + πn/2, n Є Z; x = ±2π/3 + 2πk, k Є Z.

The ability and skill to solve trigonometric equations is very important, their development requires significant effort, both on the part of the student and on the part of the teacher.

Many problems of stereometry, physics, etc. are associated with the solution of trigonometric equations. The process of solving such problems embodies many of the knowledge and skills that are acquired by studying the elements of trigonometry.

Trigonometric equations occupy an important place in the process of learning mathematics and personal development in general.

Still have questions? Don't know how to solve trigonometric equations?
To get help from a tutor, register.
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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 of mixed type. Such equations require an 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 indicated by the short entry:

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 an infinite set. To get all the other roots, it is enough to add any number of full revolutions to π /3 ( n ) in radians. Those. 2π n 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 the cosine (sine, tangent, cotangent) from the given equation on a circle, draw the angles corresponding to it 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 turns and, with a clear conscience, 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. There is not a single tricky spell about “inverse trigonometric functions” on this link... 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! The general principle is just that! 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 the selection of roots on a given interval, in trigonometric inequalities - they are generally solved 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 an outdated 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.

The video course “Get an A” includes all the topics necessary to successfully pass the Unified State Exam in mathematics with 60-65 points. Completely all tasks 1-13 of the Profile Unified State Exam in mathematics. Also suitable for passing the Basic Unified State Examination in mathematics. If you want to pass the Unified State Exam with 90-100 points, you need to solve part 1 in 30 minutes and without mistakes!

Preparation course for the Unified State Exam for grades 10-11, as well as for teachers. Everything you need to solve Part 1 of the Unified State Exam in mathematics (the first 12 problems) and Problem 13 (trigonometry). And this is more than 70 points on the Unified State Exam, and neither a 100-point student nor a humanities student can do without them.

All the necessary theory. Quick solutions, pitfalls and secrets of the Unified State Exam. All current tasks of part 1 from the FIPI Task Bank have been analyzed. The course fully complies with the requirements of the Unified State Exam 2018.

The course contains 5 large topics, 2.5 hours each. Each topic is given from scratch, simply and clearly.

Hundreds of Unified State Exam tasks. Word problems and probability theory. Simple and easy to remember algorithms for solving problems. Geometry. Theory, reference material, analysis of all types of Unified State Examination tasks. Stereometry. Tricky solutions, useful cheat sheets, development of spatial imagination. Trigonometry from scratch to problem 13. Understanding instead of cramming. Clear explanations of complex concepts. Algebra. Roots, powers and logarithms, function and derivative. A basis for solving complex problems of Part 2 of the Unified State Exam.

A lesson in the integrated application of knowledge.

Lesson objectives.

1. Review various methods for solving trigonometric equations.
2. Developing students' creative abilities by solving equations.
3. Encouraging students to self-control, mutual control, and self-analysis of their educational activities.

Equipment: screen, projector, reference material.

During the classes

Introductory conversation.

The main method for solving trigonometric equations is to reduce them to their simplest form. In this case, the usual methods are used, for example, factorization, as well as techniques used only for solving trigonometric equations. There are quite a lot of these techniques, for example, various trigonometric substitutions, angle transformations, transformations of trigonometric functions. The indiscriminate application of any trigonometric transformations usually does not simplify the equation, but catastrophically complicates it. In order to develop a general plan for solving the equation, to outline a way to reduce the equation to the simplest, you must first analyze the angles - the arguments of the trigonometric functions included in the equation.

Today we will talk about methods for solving trigonometric equations. The correctly chosen method can often significantly simplify the solution, so all the methods we have studied should always be kept in mind in order to solve trigonometric equations using the most appropriate method.

II. (Using a projector, we repeat the methods for solving equations.)

1. Method of reducing a trigonometric equation to an algebraic one.

It is necessary to express all trigonometric functions through one, with the same argument. This can be done using the basic trigonometric identity and its consequences. We obtain an equation with one trigonometric function. Taking it as a new unknown, we obtain an algebraic equation. We find its roots and return to the old unknown, solving the simplest trigonometric equations.

2. Factorization method.

To change angles, formulas for reduction, sum and difference of arguments are often useful, as well as formulas for converting the sum (difference) of trigonometric functions into a product and vice versa.

sin x + sin 3x = sin 2x + sin4x

3. Method of introducing an additional angle.

4. Method of using universal substitution.

Equations of the form F(sinx, cosx, tanx) = 0 are reduced to algebraic using a universal trigonometric substitution

Expressing sine, cosine and tangent in terms of the tangent of a half angle. This technique can lead to a higher order equation. The solution to which is difficult.