Find a solution to a trigonometric equation on a segment. Trigonometric equations - formulas, solutions, examples

Trigonometric equations- the topic is not the simplest. They are too diverse.) For example, these:

sin 2 x + cos3x = ctg5x

sin(5x+π /4) = cot(2x-π /3)

sinx + cos2x + tg3x = ctg4x

And the like...

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. Otherwise, no 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

and the like. 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. First way: using logic and 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 ones 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. If we talk human language, need to 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

Such full revolutions you can screw it up infinite set... 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 to ( ∈ ) 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. 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 it 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! He equal to angle X , just postponed to 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 wrote down these angles in 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 note that the number 0.5 is not the only one possible number in equations!) It’s just easier 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? It's 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 it. 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 obliged. Now let's expand our capabilities to all other values. Decide, so decide!)

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

Such a cosine value in brief tables No. 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. I wish I knew why equal to x, the answer would have been written down right away! We don’t know... Failure!? Calm! Mathematics does not leave its own people in trouble! She came up with arc cosines for this case. Don't 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 arc cosine in essence, no different from the picture for the equation cosx = 0.5.

That's right! 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 question!

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 infinite number not 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 most simple definitions. But remember no table values No need!)

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.

Lesson and presentation on the topic: "Solving simple trigonometric equations"

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What we will 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 arcsine, arccosine, arctangent and arccotangent. Now let's look at trigonometric equations in general.

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

Let us repeat the form of solving the simplest trigonometric equations:

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

X= ± arccos(a) + 2πk

2) If |a|≤ 1, then sin equation(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) 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: T(kx+m)=a, T is some trigonometric function.

Solve the equations: a) sin(3x)= √3/2

Solution:

A) Let us 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 return 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 let’s move directly to calculating 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 it in the form: 3x- π/3=arctg(√3)+ πk. We know that: arctan(√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 the equations: cos(4x)= √2/2. And find all the roots on the segment.

Solution:

We'll decide in general view our equation: 4x= ± arccos(√2/2) + 2πk

4x= ± π/4 + 2πk;

X= ± π/16+ πk/2;

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

Answer: x= π/16, x= 9π/16

Two main solution methods.

Let's solve the equation:

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

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

Let's find the roots quadratic equation: t=-1 and t=1/3

Then tg(x)=-1 and tg(x)=1/3, we get 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 will take the form: 2-2cos 2 (x) + 3 cos (x) = 0

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

Let us introduce the replacement t=cos(x): 2t 2 -3t - 2 = 0

The solution to our quadratic equation is 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, divide it by cos(x): You cannot divide by cosine if it equal to zero, let's make sure that this is not the case:
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 get a contradiction, so we can safely divide by zero.

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

Solution:

We'll take it out common multiplier: 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 at 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, always follow these rules!

1. See what the coefficient a is equal to, if a=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 sides of the equation by the cosine squared, we get:

We change the variable t=tg(x) and get the equation:

Solve example no.:3

Let's divide both sides of the equation by the cosine square:

We change the variable t=tg(x): t 2 + 2 t - 3 = 0

Let's 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 No.: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 no.:5

Solution:
Let's transform our expression:

Let us 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

Problems for independent solution.

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

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

3) Solve the equation: cot 2 (x) + 2 cot (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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An equality containing an unknown under the sign of a trigonometric function (`sin x, cos x, tan x` or `ctg x`) is called a trigonometric equation, and it is their formulas that we will consider further.

The simplest equations are `sin x=a, cos x=a, tg x=a, ctg x=a`, where `x` is the angle to be found, `a` is any number. Let us write down the root formulas for each of them.

1. Equation `sin x=a`.

For `|a|>1` it has no solutions.

When `|a| \leq 1` has infinite number decisions.

Root formula: `x=(-1)^n arcsin a + \pi n, n \in Z`

2. Equation `cos x=a`

For `|a|>1` - as in the case of the sine, solutions among real numbers does not have.

When `|a| \leq 1` has an infinite number of solutions.

Root formula: `x=\pm arccos a + 2\pi n, n \in Z`

Special cases for sine and cosine in graphs.

3. Equation `tg x=a`

Has an infinite number of solutions for any values ​​of `a`.

Root formula: `x=arctg a + \pi n, n \in Z`

4. Equation `ctg x=a`

Also has an infinite number of solutions for any values ​​of `a`.

Root formula: `x=arcctg a + \pi n, n \in Z`

Formulas for the roots of trigonometric equations in the table

For sine:
For cosine:
For tangent and cotangent:
Formulas for solving equations containing inverse trigonometric functions:

Methods for solving trigonometric equations

Solving any trigonometric equation consists of two stages:

• with the help of transforming it to the simplest;
• solve the simplest equation obtained using the root formulas and tables written above.

Let's look at the main solution methods using examples.

Algebraic method.

This method involves replacing a variable and substituting it into an equality.

Example. Solve the equation: `2cos^2(x+\frac \pi 6)-3sin(\frac \pi 3 - x)+1=0`

`2cos^2(x+\frac \pi 6)-3cos(x+\frac \pi 6)+1=0`,

make a replacement: `cos(x+\frac \pi 6)=y`, then `2y^2-3y+1=0`,

we find the roots: `y_1=1, y_2=1/2`, from which two cases follow:

1. `cos(x+\frac \pi 6)=1`, `x+\frac \pi 6=2\pi n`, `x_1=-\frac \pi 6+2\pi n`.

2. `cos(x+\frac \pi 6)=1/2`, `x+\frac \pi 6=\pm arccos 1/2+2\pi n`, `x_2=\pm \frac \pi 3- \frac \pi 6+2\pi n`.

Answer: `x_1=-\frac \pi 6+2\pi n`, `x_2=\pm \frac \pi 3-\frac \pi 6+2\pi n`.

Factorization.

Example. Solve the equation: `sin x+cos x=1`.

Solution. Let's move all the terms of the equality to the left: `sin x+cos x-1=0`. Using , we transform and factorize the left-hand side:

`sin x — 2sin^2 x/2=0`,

`2sin x/2 cos x/2-2sin^2 x/2=0`,

`2sin x/2 (cos x/2-sin x/2)=0`,

1. `sin x/2 =0`, `x/2 =\pi n`, `x_1=2\pi n`.
2. `cos x/2-sin x/2=0`, `tg x/2=1`, `x/2=arctg 1+ \pi n`, `x/2=\pi/4+ \pi n` , `x_2=\pi/2+ 2\pi n`.

Answer: `x_1=2\pi n`, `x_2=\pi/2+ 2\pi n`.

Reduction to a homogeneous equation

First, you need to reduce this trigonometric equation to one of two forms:

`a sin x+b cos x=0` ( homogeneous equation first degree) or `a sin^2 x + b sin x cos x +c cos^2 x=0` (homogeneous equation of the second degree).

Then divide both parts by `cos x \ne 0` - for the first case, and by `cos^2 x \ne 0` - for the second. We obtain equations for `tg x`: `a tg x+b=0` and `a tg^2 x + b tg x +c =0`, which need to be solved using known methods.

Example. Solve the equation: `2 sin^2 x+sin x cos x - cos^2 x=1`.

Solution. Let's write it down right side, like `1=sin^2 x+cos^2 x`:

`2 sin^2 x+sin x cos x — cos^2 x=` `sin^2 x+cos^2 x`,

`2 sin^2 x+sin x cos x — cos^2 x -` ` sin^2 x — cos^2 x=0`

`sin^2 x+sin x cos x — 2 cos^2 x=0`.

This is a homogeneous trigonometric equation of the second degree, we divide its left and right sides by `cos^2 x \ne 0`, we get:

`\frac (sin^2 x)(cos^2 x)+\frac(sin x cos x)(cos^2 x) — \frac(2 cos^2 x)(cos^2 x)=0`

`tg^2 x+tg x — 2=0`. Let's introduce the replacement `tg x=t`, resulting in `t^2 + t - 2=0`. The roots of this equation are `t_1=-2` and `t_2=1`. Then:

1. `tg x=-2`, `x_1=arctg (-2)+\pi n`, `n \in Z`
2. `tg x=1`, `x=arctg 1+\pi n`, `x_2=\pi/4+\pi n`, ` n \in Z`.

Answer. `x_1=arctg (-2)+\pi n`, `n \in Z`, `x_2=\pi/4+\pi n`, `n \in Z`.

Go to half corner

Example. Solve the equation: `11 sin x - 2 cos x = 10`.

Solution. Let's apply the formulas double angle, resulting in: `22 sin (x/2) cos (x/2) -` `2 cos^2 x/2 + 2 sin^2 x/2=` `10 sin^2 x/2+10 cos^ 2 x/2`

`4 tg^2 x/2 — 11 tg x/2 +6=0`

Applying the above algebraic method, we get:

1. `tg x/2=2`, `x_1=2 arctg 2+2\pi n`, `n \in Z`,
2. `tg x/2=3/4`, `x_2=arctg 3/4+2\pi n`, `n \in Z`.

Answer. `x_1=2 arctg 2+2\pi n, n \in Z`, `x_2=arctg 3/4+2\pi n`, `n \in Z`.

Introduction of auxiliary angle

In the trigonometric equation `a sin x + b cos x =c`, where a,b,c are coefficients and x is a variable, divide both sides by `sqrt (a^2+b^2)`:

`\frac a(sqrt (a^2+b^2)) sin x +` `\frac b(sqrt (a^2+b^2)) cos x =` `\frac c(sqrt (a^2) +b^2))`.

The coefficients on the left side have the properties of sine and cosine, namely the sum of their squares is equal to 1 and their modules are not greater than 1. Let us denote them as follows: `\frac a(sqrt (a^2+b^2))=cos \varphi` , ` \frac b(sqrt (a^2+b^2)) =sin \varphi`, `\frac c(sqrt (a^2+b^2))=C`, then:

`cos \varphi sin x + sin \varphi cos x =C`.

Let's take a closer look at the following example:

Example. Solve the equation: `3 sin x+4 cos x=2`.

Solution. Divide both sides of the equality by `sqrt (3^2+4^2)`, we get:

`\frac (3 sin x) (sqrt (3^2+4^2))+` `\frac(4 cos x)(sqrt (3^2+4^2))=` `\frac 2(sqrt (3^2+4^2))`

`3/5 sin x+4/5 cos x=2/5`.

Let's denote `3/5 = cos \varphi` , `4/5=sin \varphi`. Since `sin \varphi>0`, `cos \varphi>0`, then as auxiliary angle let's take `\varphi=arcsin 4/5`. Then we write our equality in the form:

`cos \varphi sin x+sin \varphi cos x=2/5`

Applying the formula for the sum of angles for the sine, we write our equality in the following form:

`sin (x+\varphi)=2/5`,

`x+\varphi=(-1)^n arcsin 2/5+ \pi n`, `n \in Z`,

`x=(-1)^n arcsin 2/5-` `arcsin 4/5+ \pi n`, `n \in Z`.

Answer. `x=(-1)^n arcsin 2/5-` `arcsin 4/5+ \pi n`, `n \in Z`.

Fractional rational trigonometric equations

These are equalities with fractions whose numerators and denominators contain trigonometric functions.

Example. Solve the equation. `\frac (sin x)(1+cos x)=1-cos x`.

Solution. Multiply and divide the right side of the equality by `(1+cos x)`. As a result we get:

`\frac (sin x)(1+cos x)=` `\frac ((1-cos x)(1+cos x))(1+cos x)`

`\frac (sin x)(1+cos x)=` `\frac (1-cos^2 x)(1+cos x)`

`\frac (sin x)(1+cos x)=` `\frac (sin^2 x)(1+cos x)`

`\frac (sin x)(1+cos x)-` `\frac (sin^2 x)(1+cos x)=0`

`\frac (sin x-sin^2 x)(1+cos x)=0`

Considering that the denominator cannot be equal to zero, we get `1+cos x \ne 0`, `cos x \ne -1`, ` x \ne \pi+2\pi n, n \in Z`.

Let's equate the numerator of the fraction to zero: `sin x-sin^2 x=0`, `sin x(1-sin x)=0`. Then `sin x=0` or `1-sin x=0`.

1. `sin x=0`, `x=\pi n`, `n \in Z`
2. `1-sin x=0`, `sin x=-1`, `x=\pi /2+2\pi n, n \in Z`.

Given that ` x \ne \pi+2\pi n, n \in Z`, the solutions are `x=2\pi n, n \in Z` and `x=\pi /2+2\pi n` , `n \in Z`.

Answer. `x=2\pi n`, `n \in Z`, `x=\pi /2+2\pi n`, `n \in Z`.

Trigonometry, and trigonometric equations in particular, are used in almost all areas of geometry, physics, and engineering. Studying begins in the 10th grade, there are always tasks for the Unified State Exam, so try to remember all the formulas of trigonometric equations - they will definitely be useful to you!

However, you don’t even need to memorize them, the main thing is to understand the essence and be able to derive it. It's not as difficult as it seems. See for yourself by watching the video.