Solving trigonometric equations tangent cotangent. Trigonometric Equations

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

sin 2 x + cos3x = ctg5x

sin (5x + π / 4) = ctg (2x-π / 3)

sinx + cos2x + tg3x = ctg4x

Etc...

But these (and all others) trigonometric monsters have two common and obligatory characteristics. The first - you will not believe - there are trigonometric functions in the equations.) Second: all expressions with x are found inside these same functions. And only there! If x appears anywhere outside, for example, sin2x + 3x = 3, this will already be a mixed-type equation. 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 most simple trigonometric equations. Why? Yes, because the solution any trigonometric equations have two stages. At the first stage, the evil equation is reduced to a simple one by means of various 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 denotes any number. Anyone.

By the way, inside the function there may not be a pure x, but some kind of expression, such as:

cos (3x + π / 3) = 1/2

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

How to solve trigonometric equations?

Trigonometric equations can be solved in two ways. First way: using logic and trigonometric circle. We will consider this path here. The second way - using memory and formulas - will be discussed 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!)

Solving equations using the trigonometric circle.

We include elementary logic and the ability to use the trigonometric circle. Can't you !? However ... It's hard 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 the tutorials ...)

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

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

cosx = 0.5

I must find the X. In human terms, you need find the angle (x), the cosine of which is 0.5.

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

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

Now let's draw the angle that this cosine gives us. Move the mouse cursor over the drawing (or tap the picture on the tablet), and see this very corner NS.

What angle is cosine 0.5?

x = π / 3

cos 60 °= cos ( π / 3) = 0,5

Someone will chuckle skeptically, yes ... They say, was it worth the 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 experts understand that there are still a whole bunch of angles here, which also give a cosine equal to 0.5.

If you turn the movable side of the OA 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 the solution to our equation, since

You can wind an infinite number of such full turns ... And all these new angles will be solutions to our trigonometric equation. And all of them must somehow be written down in response. Everything. Otherwise, the decision does not count, yes ...)

Mathematics knows how to do this in a simple and elegant way. In one short answer, write endless set solutions. This is what it looks like for our equation:

x = π / 3 + 2π n, n ∈ Z

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

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

2π is one complete revolution in radians.

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

n ∈ Z

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

This entry means that you can take any whole n ... At least -3, at least 0, at least +55. What you want. If you plug that number into the answer, you get a specific angle that will definitely solve our harsh equation.)

Or, in other words, x = π / 3 is the only root of the 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.

Everything? No. I deliberately stretch the pleasure. To remember it better.) We received only 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 also angles that also give a cosine of 0.5!

Let's go back to our picture, which was used to write down the answer. There she is:

Hover the mouse over the picture and see another corner 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 corner NS is only put back in the negative direction. This is the corner -NS. But we have already figured out the x. π / 3 or 60 °. Therefore, we can safely write:

x 2 = - π / 3

Well, and, of course, add all the angles that are obtained through full turns:

x 2 = - π / 3 + 2π n, n ∈ Z

Now that's it.) In the trigonometric circle, we saw(who understands, of course)) all angles giving a cosine equal to 0.5. And they wrote these angles in short mathematical form. The answer produced two endless 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 the given equation, draw the angles corresponding to it 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 that obvious. Well, as I said, logic is required here.)

For example, let's analyze one more 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. 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 the following picture:

Dealing with the angle first NS 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 the full turns and, with a clear conscience, write down the first series of answers:

x 1 = π / 6 + 2π n, n ∈ Z

Half done. But now we need to define second corner ... This is more cunning 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 NS equal to the angle NS ... Only it is counted from the angle π in the negative direction. Therefore, it is red.) And for the answer we need an angle, measured correctly, from the positive OX semiaxis, 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 we are interested in (drawn in green) will be equal to:

π - x

X we know it π / 6 ... Therefore, the second corner will be:

π - π / 6 = 5π / 6

We again recall the addition of full revolutions and write down the second series of responses:

x 2 = 5π / 6 + 2π n, n ∈ Z

That's all. The 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. If, of course, you know how to draw tangent and cotangent on a trigonometric circle.

In the examples above, I used the table sine and cosine value: 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 short tables. We ignore this terrible fact in cold blood. Draw a circle, mark 2/3 on the cosine axis and draw the corresponding angles. We get this picture.

Let's figure it out, for a start, with an angle in the first quarter. If I had known what X was equal to, they would have written down the answer right away! We don't know ... Failure !? Calm! Mathematics does not abandon its own in trouble! She came up with arccosines for this case. Do not know? In vain. Find out, It's much easier than you think. Under this link, there is not a single tricky incantation about "inverse trigonometric functions" ... This is superfluous in this topic.

If you are in the know, it is enough to say to yourself: "X is the angle, the cosine of which is 2/3". And right away, purely by the definition of the arccosine, you can write:

We recall additional turns 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 recorded 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 that's all! This is the correct answer. Even easier than with table values. You don't need to remember anything.) By the way, the most attentive will notice that this picture with the solution through the inverse cosine in essence, does not differ from the picture for the equation cosx = 0.5.

Exactly! The general principle is general! I specially drew two almost identical pictures. The circle shows us the angle NS by its cosine. The table is a cosine, or not - the circle does not know. What is this angle, π / 3, or what kind of inverse cosine - that's up to us.

With sine, the same song. For example:

Draw the circle again, mark the sine equal to 1/3, draw the corners. The picture looks like this:

And again the picture is almost the same as for the equation sinx = 0.5. Again, start at the corner in the first quarter. What is x 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

We deal with the second corner. In the example with a table value of 0.5, it was:

π - x

So here it will be exactly the same! 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 an absolutely correct answer. Although it doesn't look very familiar. But it is 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 at a given interval, in trigonometric inequalities - they are generally solved almost always in a circle. In short, in any tasks that are slightly more difficult than the standard ones.

Let's apply our knowledge in practice?)

Solve trigonometric equations:

At first it's simpler, right from this lesson.

Now more difficult.

Hint: This is where you have to reflect on the circle. Personally.)

And now they are 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 is one ... And how, instead of two series of answers, write one. Yes, so that not a single root of the infinite number is lost!)

Well, quite simple ones):

sinx = 0,3

cosx = π

tgx = 1,2

ctgx = 3,7

Hint: here you need to know what is an arcsine, an arcsine? What is arc tangent, arc cotangent? 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? It 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, in trigonometry, it's like crossing the road with a blindfold. 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. Instant validation testing. Learning - with interest!)

you can get acquainted with functions and derivatives.

The concept of solving trigonometric equations.

• To solve a trigonometric equation, convert it to one or more basic trigonometric equations. Solving a trigonometric equation ultimately comes down to solving four basic trigonometric equations.

Solving basic trigonometric equations.

• There are 4 types of basic trigonometric equations:
• sin x = a; cos x = a
• tg x = a; ctg x = a
• Solving basic trigonometric equations involves looking at the different x positions on the unit circle and using a conversion table (or calculator).
• Example 1.sin x = 0.866. Using a conversion table (or calculator), you get the answer: x = π / 3. The unit circle gives another answer: 2π / 3. Remember: all trigonometric functions are periodic, that is, their values ​​are repeated. For example, the periodicity of sin x and cos x is 2πn, and the periodicity of tg x and ctg x is πn. Therefore, the answer is written as follows:
• x1 = π / 3 + 2πn; x2 = 2π / 3 + 2πn.
• Example 2.cos x = -1/2. Using a conversion table (or calculator), you get the answer: x = 2π / 3. The unit circle gives another answer: -2π / 3.
• x1 = 2π / 3 + 2π; x2 = -2π / 3 + 2π.
• Example 3.tg (x - π / 4) = 0.
• Answer: x = π / 4 + πn.
• Example 4. ctg 2x = 1.732.
• Answer: x = π / 12 + πn.

Transformations used to solve trigonometric equations.

• To transform trigonometric equations, algebraic transformations (factorization, reduction of homogeneous terms, etc.) and trigonometric identities are used.
• Example 5. Using trigonometric identities, the equation sin x + sin 2x + sin 3x = 0 is transformed into the equation 4cos x * sin (3x / 2) * cos (x / 2) = 0. Thus, you need to solve the following basic trigonometric equations: cos x = 0; sin (3x / 2) = 0; cos (x / 2) = 0.

• Finding angles from known values ​​of functions.

  • Before learning methods for solving trigonometric equations, you need to learn how to find angles from known values ​​of functions. This can be done using a conversion table or calculator.
  • Example: cos x = 0.732. The calculator will give the answer x = 42.95 degrees. The unit circle will give additional angles, the cosine of which is also 0.732.

• Set the solution aside on the unit circle.

  • You can defer the solutions to the trigonometric equation on the unit circle. The solutions of the trigonometric equation on the unit circle represent the vertices of a regular polygon.
  • Example: The solutions x = π / 3 + πn / 2 on the unit circle are the vertices of a square.
  • Example: The solutions x = π / 4 + πn / 3 on the unit circle represent the vertices of a regular hexagon.

• Methods for solving trigonometric equations.

  • If a given trig equation contains only one trig function, solve that equation as the basic trig equation. If a given equation includes two or more trigonometric functions, then there are 2 methods for solving such an equation (depending on the possibility of its transformation).
    • Method 1.
  • Convert this equation to an equation of the form: f (x) * g (x) * h (x) = 0, where f (x), g (x), h (x) are the basic trigonometric equations.
  • Example 6.2cos x + sin 2x = 0. (0< x < 2π)
  • Solution. Using the double angle formula sin 2x = 2 * sin x * cos x, replace sin 2x.
  • 2cos x + 2 * sin x * cos x = 2cos x * (sin x + 1) = 0. Now solve the two basic trigonometric equations: cos x = 0 and (sin x + 1) = 0.
  • Example 7.cos x + cos 2x + cos 3x = 0. (0< x < 2π)
  • Solution: Using trigonometric identities, transform this equation into an equation of the form: cos 2x (2cos x + 1) = 0. Now solve the two basic trigonometric equations: cos 2x = 0 and (2cos x + 1) = 0.
  • Example 8.sin x - sin 3x = cos 2x. (0< x < 2π)
  • Solution: Using trigonometric identities, transform this equation into an equation of the form: -cos 2x * (2sin x + 1) = 0. Now solve the two basic trigonometric equations: cos 2x = 0 and (2sin x + 1) = 0.
    • Method 2.
  • Convert the given trigonometric equation to an equation containing only one trigonometric function. Then replace this trigonometric function with some unknown, for example, t (sin x = t; cos x = t; cos 2x = t, tg x = t; tg (x / 2) = t, etc.).
  • Example 9.3sin ^ 2 x - 2cos ^ 2 x = 4sin x + 7 (0< x < 2π).
  • Solution. In this equation, replace (cos ^ 2 x) with (1 - sin ^ 2 x) (by identity). The transformed equation is:
  • 3sin ^ 2 x - 2 + 2sin ^ 2 x - 4sin x - 7 = 0. Replace sin x with t. The equation now looks like this: 5t ^ 2 - 4t - 9 = 0. This is a quadratic equation with two roots: t1 = -1 and t2 = 9/5. The second root t2 does not satisfy the range of values ​​of the function (-1< sin x < 1). Теперь решите: t = sin х = -1; х = 3π/2.
  • Example 10.tg x + 2 tg ^ 2 x = ctg x + 2
  • Solution. Replace tg x with t. Rewrite the original equation as follows: (2t + 1) (t ^ 2 - 1) = 0. Now find t and then find x for t = tg x.

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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 we will consider their formulas further.

The simplest equations are called `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` has no solutions.

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

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

2. The equation `cos x = a`

For `| a |> 1` - as in the case of sine, it has no solutions among real numbers.

For `| 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. The equation `tg x = a`

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

Root formula: `x = arctan 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 roots of trigonometric equations in a table

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

Methods for solving trigonometric equations

The solution to any trigonometric equation consists of two stages:

• using convert it to the simplest;
• solve the obtained simplest equation using the above written root formulas and tables.

Let's look at the examples of the main methods of solving.

Algebraic method.

In this method, variable replacement and substitution into equality is done.

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`,

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

we find the roots: `y_1 = 1, y_2 = 1 / 2`, whence 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. Move all the terms of the equality to the left: `sin x + cos x-1 = 0`. Using, transform and factor the left 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 = arctan 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 bring this trigonometric equation to one of two types:

`a sin x + b cos x = 0` (homogeneous equation of the 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 by known methods.

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

Solution. Rewrite the right side as `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`. We introduce the replacement `tg x = t`, as a result,` 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 = arctan 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`.

Going to half corner

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

Solution. Apply the double angle formulas, as a result: `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 arctan 2 + 2 \ pi n`, `n \ in Z`,
2. `tg x / 2 = 3 / 4`,` x_2 = arctan 3/4 + 2 \ pi n`, `n \ in Z`.

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

Introducing an auxiliary angle

In the trigonometric equation `a sin x + b cos x = c`, where a, b, c are coefficients, and x is a variable, we 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 absolute values ​​are not greater than 1. We 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 we take `\ varphi = arcsin 4 / 5` as an auxiliary angle. 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 the 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 with trigonometric functions in the numerators and denominators.

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`.

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`.

Considering 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, engineering. The study begins in grade 10, there are definitely tasks for the exam, so try to remember all the formulas of trigonometric equations - they will definitely come in handy!

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