The simplest trigonometric equations formulas are special cases. How to solve trigonometric equations

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Lesson and presentation on the topic: "Solution of the simplest trigonometric equations"

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What will we study:
1. What are trigonometric equations?

3. Two main methods for solving trigonometric equations.
4. Homogeneous trigonometric equations.
5. Examples.

What are trigonometric equations?

Guys, we have already studied the arcsine, arccosine, arctangent and arccotangent. Now let's look at trigonometric equations in general.

Trigonometric equations - equations in which the variable is contained under the sign of the trigonometric function.

We repeat the form of solving the simplest trigonometric equations:

1) If |а|≤ 1, then the equation cos(x) = a has a solution:

X= ± arccos(a) + 2πk

2) If |а|≤ 1, then the equation sin(x) = a has a solution:

3) If |a| > 1, then the equation sin(x) = a and cos(x) = a have no solutions 4) The equation tg(x)=a has a solution: x=arctg(a)+ πk

5) The equation ctg(x)=a has a solution: x=arcctg(a)+ πk

For all formulas, k is an integer

The simplest trigonometric equations have the form: Т(kx+m)=a, T- any trigonometric function.

Solve equations: a) sin(3x)= √3/2

Solution:

A) Let's denote 3x=t, then we will rewrite our equation in the form:

The solution to this equation will be: t=((-1)^n)arcsin(√3/2)+ πn.

From the table of values ​​we get: t=((-1)^n)×π/3+ πn.

Let's go back to our variable: 3x =((-1)^n)×π/3+ πn,

Then x= ((-1)^n)×π/9+ πn/3

Answer: x= ((-1)^n)×π/9+ πn/3, where n is an integer. (-1)^n - minus one to the power of n.

More examples of trigonometric equations.

Solution:

A) This time we will go directly to the calculation of the roots of the equation right away:

X/5= ± arccos(1) + 2πk. Then x/5= πk => x=5πk

Answer: x=5πk, where k is an integer.

B) We write in the form: 3x- π/3=arctg(√3)+ πk. We know that: arctg(√3)= π/3

3x- π/3= π/3+ πk => 3x=2π/3 + πk => x=2π/9 + πk/3

Answer: x=2π/9 + πk/3, where k is an integer.

Solve equations: cos(4x)= √2/2. And find all the roots on the segment .

Solution:

Let's solve our equation in general form: 4x= ± arccos(√2/2) + 2πk

4x= ± π/4 + 2πk;

X= ± π/16+ πk/2;

Now let's see what roots fall on our segment. For k For k=0, x= π/16, we are in the given segment .
With k=1, x= π/16+ π/2=9π/16, they hit again.
For k=2, x= π/16+ π=17π/16, but here we didn’t hit, which means we won’t hit for large k either.

Answer: x= π/16, x= 9π/16

Two main solution methods.

Let's solve the equation:

Solution:
To solve our equation, we use the method of introducing a new variable, denoted: t=tg(x).

As a result of the replacement, we get: t 2 + 2t -1 = 0

Find the roots of the quadratic equation: t=-1 and t=1/3

Then tg(x)=-1 and tg(x)=1/3, we got the simplest trigonometric equation, let's find its roots.

X=arctg(-1) +πk= -π/4+πk; x=arctg(1/3) + πk.

Answer: x= -π/4+πk; x=arctg(1/3) + πk.

An example of solving an equation

Solve equations: 2sin 2 (x) + 3 cos(x) = 0

Solution:

Let's use the identity: sin 2 (x) + cos 2 (x)=1

Our equation becomes: 2-2cos 2 (x) + 3 cos (x) = 0

2 cos 2 (x) - 3 cos(x) -2 = 0

Let's introduce the replacement t=cos(x): 2t 2 -3t - 2 = 0

The solution to our quadratic equation are the roots: t=2 and t=-1/2

Then cos(x)=2 and cos(x)=-1/2.

Because cosine cannot take values ​​greater than one, then cos(x)=2 has no roots.

For cos(x)=-1/2: x= ± arccos(-1/2) + 2πk; x= ±2π/3 + 2πk

Answer: x= ±2π/3 + 2πk

Homogeneous trigonometric equations.

Equations of the form

homogeneous trigonometric equations of the second degree.

To solve a homogeneous trigonometric equation of the first degree, we divide it by cos(x): It is impossible to divide by cosine if it is equal to zero, let's make sure that this is not so:
Let cos(x)=0, then asin(x)+0=0 => sin(x)=0, but sine and cosine are not equal to zero at the same time, we got a contradiction, so we can safely divide by zero.

Solve the equation:
Example: cos 2 (x) + sin(x) cos(x) = 0

Solution:

Take out the common factor: cos(x)(c0s(x) + sin (x)) = 0

Then we need to solve two equations:

cos(x)=0 and cos(x)+sin(x)=0

Cos(x)=0 for x= π/2 + πk;

Consider the equation cos(x)+sin(x)=0 Divide our equation by cos(x):

1+tg(x)=0 => tg(x)=-1 => x=arctg(-1) +πk= -π/4+πk

Answer: x= π/2 + πk and x= -π/4+πk

How to solve homogeneous trigonometric equations of the second degree?
Guys, stick to these rules always!

1. See what the coefficient a is equal to, if a \u003d 0 then our equation will take the form cos (x) (bsin (x) + ccos (x)), an example of the solution of which is on the previous slide

2. If a≠0, then you need to divide both parts of the equation by the squared cosine, we get:

We make the change of variable t=tg(x) we get the equation:

Solve Example #:3

Divide both sides of the equation by cosine square:

We make a change of variable t=tg(x): t 2 + 2 t - 3 = 0

Find the roots of the quadratic equation: t=-3 and t=1

Then: tg(x)=-3 => x=arctg(-3) + πk=-arctg(3) + πk

Tg(x)=1 => x= π/4+ πk

Answer: x=-arctg(3) + πk and x= π/4+ πk

Solve Example #:4

Solution:
Let's transform our expression:

We can solve such equations: x= - π/4 + 2πk and x=5π/4 + 2πk

Answer: x= - π/4 + 2πk and x=5π/4 + 2πk

Solve Example #:5

Solution:
Let's transform our expression:

We introduce the replacement tg(2x)=t:2 2 - 5t + 2 = 0

The solution to our quadratic equation will be the roots: t=-2 and t=1/2

Then we get: tg(2x)=-2 and tg(2x)=1/2
2x=-arctg(2)+ πk => x=-arctg(2)/2 + πk/2

2x= arctg(1/2) + πk => x=arctg(1/2)/2+ πk/2

Answer: x=-arctg(2)/2 + πk/2 and x=arctg(1/2)/2+ πk/2

Tasks for independent solution.

A) sin(7x)= 1/2 b) cos(3x)= √3/2 c) cos(-x) = -1 d) tg(4x) = √3 e) ctg(0.5x) = -1.7

2) Solve equations: sin(3x)= √3/2. And find all the roots on the segment [π/2; π].

3) Solve the equation: ctg 2 (x) + 2ctg(x) + 1 =0

4) Solve the equation: 3 sin 2 (x) + √3sin (x) cos(x) = 0

5) Solve the equation: 3sin 2 (3x) + 10 sin(3x)cos(3x) + 3 cos 2 (3x) =0

6) Solve the equation: cos 2 (2x) -1 - cos(x) =√3/2 -sin 2 (2x)

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Trigonometric equations are not the easiest topic. Painfully they are diverse.) For example, these:

sin2x + cos3x = ctg5x

sin(5x+π /4) = ctg(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 within these same functions. And only there! If x appears somewhere outside, for example, sin2x + 3x = 3, this will be a mixed type equation. Such equations require an individual approach. Here we will not consider them.

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

So, if you have problems in 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 the function there may be not a pure x, but some kind of expression, such as:

cos(3x+π /3) = 1/2

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

How to solve trigonometric equations?

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

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

We solve equations using a trigonometric circle.

We include elementary logic and the ability to use a trigonometric circle. Can't you!? However... It will be difficult for you in trigonometry...) But it doesn't matter. Take a look at the lessons "Trigonometric circle ...... What is it?" and "Counting angles on a trigonometric circle." Everything is simple there. Unlike textbooks...)

Ah, you know!? And even mastered "Practical work with a trigonometric circle"!? Accept congratulations. This topic will be close and understandable to you.) What is especially pleasing is that the trigonometric circle does not care which equation you solve. Sine, cosine, tangent, cotangent - everything is the same for him. The solution principle is the same.

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

cosx = 0.5

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

How did we use the circle before? We drew a corner on it. In degrees or radians. And immediately seen trigonometric functions of this angle. Now let's do the opposite. Draw a cosine equal to 0.5 on the circle and immediately we'll see corner. It remains only to write down the answer.) Yes, yes!

We 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 a tablet), and see this same corner X.

Which angle has a cosine of 0.5?

x \u003d π / 3

cos 60°= cos( π /3) = 0,5

Some people will grunt skeptically, yes... They say, was it worth it to fence the circle, when everything is clear anyway... You can, of course, grunt...) But the fact is that this is an erroneous answer. Or rather, inadequate. Connoisseurs of the circle understand that there are still a whole bunch of angles that also give a cosine equal to 0.5.

If you turn the movable side OA for a full turn, point A will return to its original position. With the same cosine equal to 0.5. Those. the angle will change 360° or 2π radians, and cosine is not. The new angle 60° + 360° = 420° will also be a solution to our equation, because

There are an infinite number of such full rotations... And all these new angles will be solutions to our trigonometric equation. And they all need to be written down somehow. All. Otherwise, the decision is not considered, yes ...)

Mathematics can do this simply and elegantly. In one short answer, write down infinite set solutions. Here's what it looks like for our equation:

x = π /3 + 2π n, n ∈ Z

I will decipher. Still write meaningfully nicer than stupidly drawing some mysterious letters, right?)

π /3 is the same angle that we saw on the circle and identified according to the table of cosines.

2π is one full turn in radians.

n - this is the number of complete, i.e. whole revolutions. It is clear that n can be 0, ±1, ±2, ±3.... and so on. As indicated by the short entry:

n ∈ Z

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

This notation means that you can take any integer n . At least -3, at least 0, at least +55. What do you want. If you plug that number into your answer, you get a specific angle, which is sure to be the solution to our harsh equation.)

Or, in other words, x \u003d 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 turns to π / 3 ( n ) in radians. Those. 2πn radian.

Everything? No. I specifically stretch the pleasure. To remember better.) We received only a part of the answers to our equation. I will write this first part of the solution as follows:

x 1 = π /3 + 2π n, n ∈ Z

x 1 - not one root, it is a whole series of roots, written in short form.

But there are other angles that also give a cosine equal to 0.5!

Let's return to our picture, according to which we wrote down the answer. There she is:

Move the mouse over the image and see another corner that also gives a cosine of 0.5. What do you think it equals? The triangles are the same... Yes! It is equal to the angle X , only plotted 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 \u003d - π / 3

And, of course, we add all the angles that are obtained through full turns:

x 2 = - π /3 + 2π n, n ∈ Z

That's all now.) In a trigonometric circle, we saw(who understands, of course)) all angles that give a cosine equal to 0.5. And they wrote down these angles in a short mathematical form. The answer is 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 with the help of a circle is understandable. We mark the cosine (sine, tangent, cotangent) from the given equation on the circle, draw the corresponding angles and write down the answer. Of course, you need to figure out what kind of corners we are saw on the circle. Sometimes it's not so obvious. Well, as I said, logic is required here.)

For example, let's analyze another trigonometric equation:

Please note that the number 0.5 is not the only possible number in the 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 at once all the angles corresponding to this sine. 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. The matter is simple:

x \u003d π / 6

We recall 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. Now we need to define second corner... This is trickier than in 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 the 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 semiaxis OX, i.e. from an angle of 0 degrees.

Hover the cursor over the picture and see everything. I removed the first corner so as not to complicate the picture. The angle of interest to us (drawn in green) will be equal to:

π - x

x we know it π /6 . So the second angle will be:

π - π /6 = 5π /6

Again, we recall the addition of 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

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

In the examples above, I used the tabular 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 the following trigonometric equation:

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

We understand, for starters, with an angle in the first quarter. To know what x is equal to, they would immediately write down the answer! We don't know... Failure!? Calm! Mathematics does not leave its own in trouble! She invented arc cosines for this case. Do not know? In vain. Find out. It's a lot easier than you think. According to this link, there is not a single tricky spell about "inverse trigonometric functions" ... It's superfluous in this topic.

If you're in the know, just say to yourself, "X is an angle whose cosine is 2/3." And immediately, purely by definition of the arccosine, we can write:

We remember about 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 is also written almost automatically, for the second angle. Everything is the same, only x (arccos 2/3) will be with a minus:

x 2 = - arccos 2/3 + 2π n, n ∈ Z

And all things! This is the correct answer. Even easier than with tabular values. You don’t need to remember anything.) By the way, the most attentive will notice that this picture with the solution through the arc cosine is essentially no different from the picture for the equation cosx = 0.5.

Exactly! The general principle on that and the general! I specifically drew two almost identical pictures. The circle shows us the angle X by its cosine. It is a tabular cosine, or not - the circle does not know. What kind of angle is this, π / 3, or what kind of arc cosine is up to us to decide.

With a sine the same song. For example:

Again we draw a circle, mark the sine equal to 1/3, draw the corners. It turns out this picture:

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!

So the first pack of roots is ready:

x 1 = arcsin 1/3 + 2π n, n ∈ Z

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

π - x

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

x 2 = π - arcsin 1/3 + 2π n, n ∈ Z

This is a completely correct answer. Although it does not look very familiar. But it's understandable, 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 complicated than standard ones.

Putting knowledge into practice?

Solve trigonometric equations:

At first it is simpler, directly on this lesson.

Now it's more difficult.

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

And now outwardly unpretentious ... 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 down one instead of two series of answers. Yes, so that not a single root from an infinite number is lost!)

Well, quite simple):

sinx = 0,3

cosx = π

tgx = 1,2

ctgx = 3,7

Hint: here you need to know what is the arcsine, arccosine? What is arc tangent, arc tangent? The simplest definitions. But you don’t need to remember any tabular values!)

The answers are, of course, in disarray):

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

Not everything works out? It happens. Read the lesson again. Only thoughtfully(there is such an obsolete word...) And follow the links. The main links are about the circle. Without it in trigonometry - how to cross 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. Learning - with interest!)

you can get acquainted with functions and derivatives.

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

Factoring 8

Reduction to a 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, as a rule, is unambiguously defined. For example, linear and quadratic equations and inequalities, fractional equations and equations reducible to quadratic ones, etc. Without analyzing in detail the principle of solving each of the examples mentioned, we note the general thing that is necessary for their successful solution.

In most cases, you need to determine what type of task is, remember the sequence of actions leading to the goal, and perform these actions. It is obvious that the success or failure of the student in mastering the methods of solving equations depends mainly on how much he will be able to correctly determine the type of equation and remember the sequence of all stages of its solution. Of course, this assumes that the student has the skills to perform identical transformations and calculations.

A completely different situation occurs when a student encounters trigonometric equations. At the same time, 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 choose the desired formula from the several dozen available.

To help students find their way through the complex labyrinth of trigonometric equations, they are first introduced to the equations, which, after the introduction of a new variable, are reduced to square ones. Then solve homogeneous equations and reduced to them. Everything ends, as a rule, with equations, for the solution of which it is necessary to factorize the left side, then equating each of the factors to zero.

Realizing that the one and a half dozen equations analyzed in the lessons are clearly not enough to let the student sail independently on the trigonometric "sea", the teacher adds a few more recommendations from himself.

To solve the trigonometric equation, we must try:

Bring all the functions included in the equation to "the same angles";

Bring the equation to "the same functions";

Factorize the left side of the equation, etc.

But, despite the knowledge of the main types of trigonometric equations and several principles for finding their solution, many students still find themselves at an impasse in front of each equation that differs slightly from those that were solved before. It remains unclear what one should strive for, having one or another equation, why in one case it is necessary to apply the double angle formulas, in the other - the half angle, and in the third - the 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 the same angles if all the trigonometric functions included in it have equal arguments. A trigonometric equation is said to have the same functions if it contains only one of the trigonometric functions.

Definition 3. The degree 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 the monomials 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.

The solution of trigonometric equations consists of two stages: the transformation of the equation to obtain its simplest form and the solution of 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 replacement of variables and substitution).

Solve equations.

1)

Let's introduce the notation x=2 sin3 t, we get

Solving this equation, we get:
or

those. can be written

When writing the solution obtained due to the presence of signs degree
there is no point in writing.

Answer:

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:
;
.

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.

Denoting
, we get
Solving this quadratic equation, we have:

does not satisfy the condition

We write down the solution of 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. Solution of equations using the condition of equality of the trigonometric functions of the same name.

a)
, if

b)
, if

in)
, if

Using these conditions, consider the solution of the following equations:

6)

Using what was said in item 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 part 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: .

We transform and factorize the expression on the left side of the equation:
.

.

.

1)
2)

Because
and
do not take the value null

at the same time, then we separate 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;

Put all common factors out of brackets;

Equate all factors and brackets to zero;

Parentheses equated 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
.

Consider examples:

12) Solve the equation:

Solution.

Divide both sides of the equation by
,

Introducing the notation
, name

the roots of this equation are:

from here 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 like terms, we have:

Dividing the homogeneous last equation by
, we get

I will designate
, we get the quadratic equation
, whose roots are numbers

In this way

Expression
vanishes at
, i.e. at
,
.

Our solution to the equation 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.

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 label them accordingly
(here - auxiliary angle) and our equation takes the form: .

Then

And his decision

Note that the introduced notation is 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 an equation of the form has 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

We transform the sum of trigonometric functions into a product:

Answer:

VI. Convert product to sum.

The corresponding formulas are used here.

17) Solve the equation:

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

VII.Universal substitution.

,

these formulas are true for all

Substitution
called universal.

18) Solve the equation:

Solution: Replace and
to their expression through
and denote
.

We get a rational equation
, which is converted 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 to the original equation.

Answer:
.

Comment. Equation 18 could be solved in a different way.

Divide both sides of this equation by 5 (i.e. by
):
.

Because
, then there is a number
, what
and
. So the equation becomes:
or
. From here we find that
where
.

19) Solve the equation
.

Solution. Since the functions
and
have the largest value equal to 1, then their sum is equal to 2 if
and
, at the same time, that is
.

Answer:
.

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

Conclusion.

Working on the topic “Solutions of 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 the analysis of the equation and the signs of the expediency of using one or another solution method.

To think over ways of self-control of the activity on implementation of the method.

Learn to make "your" equations for each of the studied methods.

Application No. 1

Solve homogeneous or reducible equations.