Task No. 1

The logic is simple: we will do as we did before, regardless of the fact that now trigonometric functions have a more complex argument!

If we were to solve an equation of the form:

Then we would write down the following answer:

Or (since)

But now our role is played by this expression:

Then we can write:

Our goal with you is to make sure that the left side stands simply, without any “impurities”!

Let's gradually get rid of them!

First, let’s remove the denominator at: to do this, multiply our equality by:

Now let's get rid of it by dividing both parts:

Now let's get rid of the eight:

The resulting expression can be written as 2 series of solutions (by analogy with a quadratic equation, where we either add or subtract the discriminant)

We need to find the largest negative root! It is clear that we need to sort through.

Let's look at the first episode first:

It is clear that if we take, then as a result we will receive positive numbers, but they do not interest us.

So you need to take it negative. Let be.

When the root will be narrower:

And we need to find the greatest negative!! This means that going in the negative direction no longer makes sense here. And the largest negative root for this series will be equal to.

Now let's look at the second series:

And again we substitute: , then:

Not interested!

Then it makes no sense to increase any more! Let's reduce it! Let then:

Fits!

Let be. Then

Then - the largest negative root!

Answer:

Task No. 2

We solve again, regardless of the complex cosine argument:

Now we express again on the left:

Multiply both sides by

Divide both sides by

All that remains is to move it to the right, changing its sign from minus to plus.

We again get 2 series of roots, one with and the other with.

We need to find the largest negative root. Let's look at the first episode:

It is clear that we will get the first negative root at, it will be equal to and will be the largest negative root in 1 series.

For the second series

The first negative root will also be obtained at and will be equal to. Since, then is the largest negative root of the equation.

Answer: .

Task No. 3

We solve, regardless of the complex tangent argument.

Now, it doesn’t seem complicated, right?

As before, we express on the left side:

Well, that’s great, there’s only one series of roots here! Let's find the largest negative again.

It is clear that it turns out if you put it down. And this root is equal.

Answer:

Now try to solve the following problems yourself.

Homework or 3 tasks to solve independently.

1. Resolve the equation.
   
2. Resolve the equation.
   In the answer to the pi-shi-th-the-smallest po-lo-living root.
3. Resolve the equation.
   In the answer to the pi-shi-th-the-smallest po-lo-living root.

Ready? Let's check. I will not describe in detail the entire solution algorithm; it seems to me that it has already received enough attention above.

Well, is everything right? Oh, those nasty sinuses, there’s always some kind of trouble with them!

Well, now you can solve simple trigonometric equations!

Check out the solutions and answers:

Task No. 1

Let's express

The smallest positive root is obtained if we put, since, then

Answer:

Task No. 2

The smallest positive root is obtained at.

It will be equal.

Answer: .

Task No. 3

When we receive, when we have.

Answer: .

This knowledge will help you solve many problems that you will encounter in the exam.

If you are applying for a “5” rating, then you just need to proceed to reading the article for mid-level, which will be devoted to solving more complex trigonometric equations (task C1).

AVERAGE LEVEL

In this article I will describe solving more complex trigonometric equations and how to select their roots. Here I will draw on the following topics:

1. Trigonometric equations for beginner level (see above).

More complex trigonometric equations are the basis for advanced problems. They require how to solve the equation itself in general view, and find the roots of this equation belonging to some given interval.

Solving trigonometric equations comes down to two subtasks:

1. Solving the equation
2. Root selection

It should be noted that the second is not always required, but in most examples selection is still required. But if it is not required, then we can sympathize with you - this means that the equation is quite complex in itself.

My experience in analyzing C1 problems shows that they are usually divided into the following categories.

Four categories of tasks of increased complexity (formerly C1)

1. Equations that reduce to factorization.
2. Equations reduced to form.
3. Equations solved by changing a variable.
4. Equations that require additional selection of roots due to irrationality or denominator.

To put it simply: if you get caught one of the equations of the first three types, then consider yourself lucky. For them, as a rule, you additionally need to select roots belonging to a certain interval.

If you come across a type 4 equation, then you are less lucky: you need to tinker with it longer and more carefully, but quite often it does not require additional selection of roots. Nevertheless, I will analyze this type of equations in the next article, and this one I will devote to solving equations of the first three types.

Equations that reduce to factorization

The most important thing you need to remember to solve this type of equation is

As practice shows, as a rule, this knowledge is sufficient. Let's look at some examples:

Example 1. Equation reduced to factorization using the reduction and double angle sine formulas

• Resolve the equation
• Find all the roots of this equation that lie above the cut

Here, as I promised, the reduction formulas work:

Then my equation will look like this:

Then my equation will take the following form:

A short-sighted student might say: now I’ll reduce both sides by, get the simplest equation and enjoy life! And he will be bitterly mistaken!

REMEMBER: YOU CAN NEVER REDUCE BOTH SIDES OF A TRIGONOMETRIC EQUATION BY A FUNCTION CONTAINING AN UNKNOWN! SO YOU LOSE YOUR ROOTS!

So what to do? Yes, it’s simple, move everything to one side and take out the common factor:

Well, we factored it into factors, hurray! Now let's decide:

The first equation has roots:

And the second:

This completes the first part of the problem. Now you need to select the roots:

The gap is like this:

Or it can also be written like this:

Well, let's take the roots:

First, let's work with the first episode (and it's simpler, to say the least!)

Since our interval is entirely negative, there is no need to take non-negative ones, they will still give non-negative roots.

Let's take it, then - it's too much, it doesn't hit.

Let it be, then - I didn’t hit it again.

One more try - then - yes, I got it! The first root has been found!

I shoot again: then I hit it again!

Well, one more time: : - this is already a flight.

So from the first series there are 2 roots belonging to the interval: .

We are working with the second series (we are building to the power according to the rule):

Undershoot!

Missing it again!

Missing it again!

Got it!

Flight!

Thus, my interval has the following roots:

This is the algorithm we will use to solve all other examples. Let's practice together with one more example.

Example 2. Equation reduced to factorization using reduction formulas

• Solve the equation

Solution:

Again the notorious reduction formulas:

Don't try to cut back again!

The first equation has roots:

And the second:

Now again the search for roots.

I’ll start with the second episode, I already know everything about it from the previous example! Look and make sure that the roots belonging to the interval are as follows:

Now the first episode and it’s simpler:

If - suitable

If that's fine too

If it’s already a flight.

Then the roots will be as follows:

Independent work. 3 equations.

Well, is the technique clear to you? Does solving trigonometric equations not seem so difficult anymore? Then quickly solve the following problems yourself, and then we will solve other examples:

1. Solve the equation
   Find all the roots of this equation that lie above the interval.
2. Resolve the equation
   Indicate the roots of the equation that lie above the cut
3. Resolve the equation
   Find all the roots of this equation that lie between them.

Equation 1.

And again the reduction formula:

First series of roots:

Second series of roots:

We begin selection for the gap

Answer: , .

Equation 2. Checking independent work.

Quite a tricky grouping into factors (I’ll use the double angle sine formula):

then or

This is a general solution. Now we need to select the roots. The trouble is that we cannot tell the exact value of an angle whose cosine is equal to one quarter. Therefore, I can’t just get rid of the arc cosine - such a shame!

What I can do is figure out that so, so, then.

Let's create a table: interval:

Well, through painful searches we came to the disappointing conclusion that our equation has one root on the indicated interval: \displaystyle arccos\frac(1)(4)-5\pi

Equation 3: Independent work test.

A frightening looking equation. However, it can be solved quite simply by applying the double angle sine formula:

Let's reduce it by 2:

Let's group the first term with the second and the third with the fourth and take out the common factors:

It is clear that the first equation has no roots, and now let’s consider the second:

In general, I was going to dwell a little later on solving such equations, but since it turned up, there’s nothing to do, I have to solve it...

Equations of the form:

This equation is solved by dividing both sides by:

Thus, our equation has a single series of roots:

We need to find those that belong to the interval: .

Let's build a table again, as I did earlier:

Answer: .

Equations reduced to the form:

Well, now is the time to move on to the second portion of equations, especially since I have already spilled the beans on what the solution to trigonometric equations of a new type consists of. But it is worth repeating that the equation is of the form

Solved by dividing both sides by cosine:

1. Resolve the equation
   Indicate the roots of the equation that lie above the cut.
2. Resolve the equation
   Indicate the roots of the equation that lie between them.

Example 1.

The first one is quite simple. Move to the right and apply the double angle cosine formula:

Yeah! Equation of the form: . I divide both parts by

We do root screening:

Gap:

Answer:

Example 2.

Everything is also quite trivial: let’s open the brackets on the right:

Basic trigonometric identity:

Sine of double angle:

Finally we get:

Root screening: interval.

Answer: .

Well, how do you like the technique, isn’t it too complicated? I hope not. We can immediately make a reservation: in their pure form, equations that immediately reduce to an equation for the tangent are quite rare. Typically, this transition (division by cosine) is only part of a more complex problem. Here's an example for you to practice:

• Resolve the equation
• Find all the roots of this equation that lie above the cut.

Let's check:

The equation can be solved immediately; it is enough to divide both sides by:

Root screening:

Answer: .

One way or another, we have yet to encounter equations of the type that we have just examined. However, it is too early for us to call it a day: there is still one more “layer” of equations that we have not analyzed. So:

Solving trigonometric equations by changing variables

Everything is transparent here: we look closely at the equation, simplify it as much as possible, make a substitution, solve it, make a reverse substitution! In words everything is very easy. Let's see in action:

Example.

• Solve the equation: .
• Find all the roots of this equation that lie above the cut.

Well, here the replacement itself suggests itself to us!

Then our equation will turn into this:

The first equation has roots:

And the second one is like this:

Now let's find the roots belonging to the interval

Answer: .

Let's look at a slightly more complex example together:

• Resolve the equation
• Indicate the roots of the given equation, lying above-lying between them.

Here the replacement is not immediately visible, moreover, it is not very obvious. Let's first think: what can we do?

We can, for example, imagine

And at the same time

Then my equation will take the form:

And now attention, focus:

Let's divide both sides of the equation by:

Suddenly you and I have a quadratic equation relative! Let's make a replacement, then we get:

The equation has the following roots:

Unpleasant second series of roots, but nothing can be done! We select roots in the interval.

We also need to consider that

Since and, then

Answer:

To reinforce this before you solve the problems yourself, here’s another exercise for you:

• Resolve the equation
• Find all the roots of this equation that lie between them.

Here you need to keep your eyes open: we now have denominators that can be zero! Therefore, you need to be especially attentive to the roots!

First of all, I need to rearrange the equation so that I can make a suitable substitution. I can’t think of anything better now than to rewrite the tangent in terms of sine and cosine:

Now I will move from cosine to sine using the basic trigonometric identity:

And finally, I’ll bring everything to a common denominator:

Now I can move on to the equation:

But at (that is, at).

Now everything is ready for replacement:

Then or

However, note that if, then at the same time!

Who suffers from this? The problem with the tangent is that it is not defined when the cosine is equal to zero (division by zero occurs).

Thus, the roots of the equation are:

Now we sift out the roots in the interval:

	- fits
	- overkill

Thus, our equation has a single root on the interval, and it is equal.

You see: the appearance of a denominator (just like the tangent, leads to certain difficulties with the roots! Here you need to be more careful!).

Well, you and I have almost finished analyzing trigonometric equations; there is very little left - to solve two problems on your own. Here they are.

1. Solve the equation
   Find all the roots of this equation that lie above the cut.
2. Resolve the equation
   Indicate the roots of this equation, located above the cut.

Decided? Isn't it very difficult? Let's check:

1. We work according to the reduction formulas:

   Substitute into the equation:

   Let's rewrite everything through cosines to make it easier to make the replacement:

   Now it's easy to make a replacement:

   It is clear that is an extraneous root, since the equation has no solutions. Then:

   We are looking for the roots we need in the interval

   Answer: .

2. 
   Here the replacement is immediately visible:

   Then or

   	- fits!	- fits!
   	- fits!	- fits!
   	- a lot of!	- also a lot!

   Answer:

Well, that's it now! But solving trigonometric equations does not end there; we are left behind in the most difficult cases: when the equations contain irrationality or various kinds of “complex denominators.” We will look at how to solve such tasks in an article for an advanced level.

ADVANCED LEVEL

In addition to the trigonometric equations discussed in the previous two articles, we will consider another class of equations that require even more careful analysis. These trigonometric examples contain either irrationality or a denominator, which makes their analysis more difficult. However, you may well encounter these equations in Part C of the exam paper. However, every cloud has a silver lining: for such equations, as a rule, the question of which of its roots belongs to a given interval is no longer raised. Let's not beat around the bush, but let's go straight to trigonometric examples.

Example 1.

Solve the equation and find the roots that belong to the segment.

Solution:

We have a denominator that should not be equal to zero! Then solving this equation is the same as solving the system

Let's solve each of the equations:

And now the second one:

Now let's look at the series:

It is clear that this option does not suit us, since in this case our denominator is reset to zero (see the formula for the roots of the second equation)

If, then everything is in order, and the denominator is not zero! Then the roots of the equation are as follows: , .

Now we select the roots belonging to the interval.

	- not suitable	- fits
	- fits	- fits
	overkill	overkill

Then the roots are as follows:

You see, even the appearance of a small disturbance in the form of the denominator significantly affected the solution of the equation: we discarded a series of roots that nullified the denominator. Things can get even more complicated if you come across trigonometric examples that are irrational.

Example 2.

Solve the equation:

Solution:

Well, at least you don’t have to take away the roots, and that’s good! Let's first solve the equation, regardless of irrationality:

So, is that all? No, alas, it would be too easy! We must remember that only non-negative numbers can appear under the root. Then:

The solution to this inequality is:

Now it remains to find out whether part of the roots of the first equation inadvertently ended up where the inequality does not hold.

To do this, you can again use the table:

	: , But	No!
		Yes!
		Yes!

Thus, one of my roots “fell out”! It turns out if you put it down. Then the answer can be written as follows:

Answer:

You see, the root requires even more attention! Let's make it more complicated: let now I have a trigonometric function under my root.

Example 3.

As before: first we will solve each one separately, and then we will think about what we have done.

Now the second equation:

Now the most difficult thing is to find out whether negative values ​​​​are obtained under the arithmetic root if we substitute the roots from the first equation there:

The number must be understood as radians. Since a radian is approximately degrees, then radians are on the order of degrees. This is the corner of the second quarter. What is the sign of the cosine of the second quarter? Minus. What about sine? Plus. So what can we say about the expression:

It's less than zero!

This means that it is not the root of the equation.

Now it's time.

Let's compare this number with zero.

Cotangent is a function decreasing in 1 quarter (the smaller the argument, the greater the cotangent). radians are approximately degrees. In the same time

since, then, and therefore
,

Answer: .

Could it get any more complicated? Please! It will be more difficult if the root is still a trigonometric function, and the second part of the equation is again a trigonometric function.

The more trigonometric examples the better, see below:

Example 4.

The root is not suitable due to the limited cosine

Now the second one:

At the same time, by definition of a root:

We need to remember the unit circle: namely, those quarters where the sine is less than zero. What are these quarters? Third and fourth. Then we will be interested in those solutions of the first equation that lie in the third or fourth quarter.

The first series gives roots lying at the intersection of the third and fourth quarters. The second series - diametrically opposite to it - gives rise to roots lying on the border of the first and second quarters. Therefore, this series is not suitable for us.

Answer: ,

And again trigonometric examples with "difficult irrationality". Not only do we have the trigonometric function under the root again, but now it is also in the denominator!

Example 5.

Well, nothing can be done - we do as before.

Now we work with the denominator:

I don’t want to solve the trigonometric inequality, so I’ll do something cunning: I’ll take and substitute my series of roots into the inequality:

If - is even, then we have:

since all angles of the view lie in the fourth quarter. And again the sacred question: what is the sign of the sine in the fourth quarter? Negative. Then the inequality

If -odd, then:

In which quarter does the angle lie? This is the corner of the second quarter. Then all the corners are again the corners of the second quarter. The sine there is positive. Just what you need! So the series:

Fits!

We deal with the second series of roots in the same way:

We substitute into our inequality:

If - even, then

First quarter corners. The sine there is positive, which means the series is suitable. Now if - odd, then:

fits too!

Well, now we write down the answer!

Answer:

Well, this was perhaps the most labor-intensive case. Now I offer you problems to solve on your own.

Training

1. Solve and find all the roots of the equation that belong to the segment.

Solutions:

1. 
   First equation:
   or
   ODZ of the root:

   Second equation:

   Selection of roots that belong to the interval

   Answer:

Or
or
But

Let's consider: . If - even, then
- doesn't fit!
If - odd, : - suitable!
This means that our equation has the following series of roots:
or
Selection of roots in the interval:

	- not suitable	- fits
	- fits	- a lot of
	- fits	a lot of

Answer: , .

Or
Since, then the tangent is not defined. We immediately discard this series of roots!

Second part:

At the same time, according to DZ it is required that

We check the roots found in the first equation:

If the sign:

First quarter angles where the tangent is positive. Doesn't fit!
If the sign:

Fourth quarter corner. There the tangent is negative. Fits. We write down the answer:

Answer: , .

We have looked at complex trigonometric examples together in this article, but you should solve the equations yourself.

SUMMARY AND BASIC FORMULAS

A trigonometric equation is an equation in which the unknown is strictly under the sign of the trigonometric function.

There are two ways to solve trigonometric equations:

The first way is using formulas.

The second way is through the trigonometric circle.

Allows you to measure angles, find their sines, cosines, etc.

The purpose of the lesson:

A) strengthen the ability to solve simple trigonometric equations;

b) teach how to select roots of trigonometric equations from a given interval

During the classes.

1. Updating knowledge.

a)Checking homework: the class is given advanced homework– solve the equation and find a way to select roots from a given interval.

1)cos x= -0.5, where xI [- ]. Answer:.

2) sin x= , where xI . Answer: ; .

3)cos 2 x= -, where xI. Answer:

Students write down the solution on the board, some using a graph, others using the selection method.

At this time class works orally.

Find the meaning of the expression:

a) tg – sin + cos + sin. Answer: 1.

b) 2arccos 0 + 3 arccos 1. Answer: ?

c) arcsin + arcsin. Answer:.

d) 5 arctg (-) – arccos (-). Answer:-.

– Let’s check your homework, open your notebooks with homework.

Some of you found the solution using the selection method, and some using the graph.

2. Conclusion about ways to solve these tasks and statement of the problem, i.e., communication of the topic and purpose of the lesson.

– a) It is difficult to solve using selection if a large interval is given.

– b) The graphical method does not give accurate results, requires verification, and takes a lot of time.

– Therefore, there must be at least one more method, the most universal one - let’s try to find it. So, what are we going to do in class today? (Learn to choose the roots of a trigonometric equation on a given interval.)

– Example 1. (Student goes to the board)

cos x= -0.5, where xI [- ].

Question: What determines the answer to this task? (From the general solution of the equation. Let's write the solution in general form). The solution is written on the board

x = + 2?k, where k R.

– Let’s write this solution in the form of a set:

– In your opinion, in what notation of the solution is it convenient to choose roots on the interval? (from the second entry). But this is again a selection method. What do we need to know to get the right answer? (You need to know the values ​​of k).

(Let's create a mathematical model to find k).

since kI Z, then k = 0, hence X= =	From this inequality it is clear that there are no integer values ​​of k.

Conclusion: To select roots from a given interval when solving a trigonometric equation, you need to:

1. to solve an equation of the form sin x = a, cos x = a It is more convenient to write the roots of the equation as two series of roots.
2. to solve equations of the form tan x = a, ctg x = a write down the general formula for roots.
3. create a mathematical model for each solution in the form of a double inequality and find the integer value of the parameter k or n.
4. substitute these values ​​into the root formula and calculate them.

3. Consolidation.

Solve example No. 2 and No. 3 from homework using the resulting algorithm. Two students work at the board at the same time, followed by checking the work.

Mandatory minimum knowledge

sin x = a, -1 a 1 (a 1)
x = arcsin a + 2 n, n Z
x = - arcsin a + 2 n, n Z
or
x = (- 1)k arcsin a + k, k Z
arcsin (- a) = - arcsin a
sin x = 1
x = /2 + 2 k, k Z
sin x = 0
x = k, k Z
sin x = - 1
x = - /2 + 2 k, k Z
y
y
x
y
x
x

Mandatory minimum knowledge

cos x = a, -1 a 1 (a 1)
x = arccos a + 2 n, n Z
arccos (- a) = - arccos a
cos x = 1
x = 2 k, k Z
cos x = 0
x = /2 + k, k Z
y
y
x
cos x = - 1
x = + 2 k, k Z
y
x
x

Mandatory minimum knowledge

tg x = a, a R
x = arctan a + n, n Z
cot x = a, a R
x = arcctg a + n, n Z
arctg (- a) = - arctg a
arctg (- a) = - arctg a Reduce the equation to one function
Reduce to one argument
Some solution methods
trigonometric equations
Application of trigonometric formulas
Using abbreviated multiplication formulas
Factorization
Reduction to quadratic equation relative to sin x, cos x, tan x
By introducing an auxiliary argument
By dividing both sides of a homogeneous equation of the first degree
(asin x +bcosx = 0) by cos x
By dividing both sides of a homogeneous equation of the second degree
(a sin2 x +bsin x cos x+ c cos2x =0) by cos2 x

Oral Exercises Calculate

arcsin ½
arcsin (- √2/2)
arccos √3/2
arccos (-1/2)
arctan √3
arctan (-√3/3)
= /6
= - /4
= /6
= - arccos ½ = - /3 = 2 /3
= /3
= - /6

(using a trigonometric circle)
cos 2x = ½, x [- /2; 3 /2]
2x = ± arccos ½ + 2 n, n Z
2x = ± /3 + 2 n, n Z
x = ± /6 + n, n Z
Let's select roots using a trigonometric circle
Answer: - /6; /6; 5 /6; 7 /6

Various methods of root selection

Find the roots of the equation belonging to the given interval
sin 3x = √3/2, x [- /2; /2]
3x = (– 1)k /3 + k, k Z
x = (– 1)k /9 + k/3, k Z
Let's select the roots by enumerating the values ​​of k:
k = 0, x = /9 – belongs to the interval
k = 1, x = – /9 + /3 = 2 /9 – belongs to the interval
k = 2, x = /9 + 2 /3 = 7 /9 – does not belong to the interval
k = – 1, x = – /9 – /3 = – 4 /9 – belongs to the interval
k = – 2, x = /9 – 2 /3 = – 5 /9 – does not belong to the interval
Answer: -4 /9; /9; 2 /9

Various methods of root selection

Find the roots of the equation belonging to the given interval
(using inequality)
tg 3x = – 1, x (- /2;)
3x = – /4 + n, n Z
x = – /12 + n/3, n Z
Let's select the roots using the inequality:
– /2 < – /12 + n/3 < ,
– 1/2 < – 1/12 + n/3 < 1,
– 1/2 + 1/12 < n/3 < 1+ 1/12,
– 5/12 < n/3 < 13/12,
– 5/4 < n < 13/4, n Z,
n = – 1; 0; 1; 2; 3
n = – 1, x = – /12 – /3 = – 5 /12
n = 0, x = – /12
n = 1, x = – /12 + /3 = /4
n = 2, x = – /12 + 2 /3 = 7 /12
n = 3, x = – /12 + = 11 /12
Answer: – 5 /12; - /12; /4; 7 /12; 11/12

10. Various methods of root selection

Find the roots of the equation belonging to the given interval
(using graph)
cos x = – √2/2, x [–4; 5 /4]
x = arccos (– √2/2) + 2 n, n Z
x = 3 /4 + 2 n, n Z
Let's select the roots using the graph:
x = – /2 – /4 = – 3 /4; x = – – /4 = – 5 /4
Answer: 5 /4; 3/4

11. 1. Solve the equation 72cosx = 49sin2x and indicate its roots on the segment [; 5/2]

1. Solve the equation 72cosx = 49sin2x
and indicate its roots on the segment [; 5 /2]
Let's solve the equation:
72cosx = 49sin2x,
72cosx = 72sin2x,
2cos x = 2sin 2x,
cos x – 2 sinx cosx = 0,
cos x (1 – 2sinx) = 0,
cos x = 0 ,
x = /2 + k, k Z
or
1 – 2sinx = 0,
sin x = ½,
x = (-1)n /6 + n, n Z
Let's select roots using
trigonometric circle:
x = 2 + /6 = 13 /6
Answer:
a) /2 + k, k Z, (-1)n /6 + n, n Z
b) 3 /2; 5 /2; 13/6

12. 2. Solve the equation 4cos2 x + 8 cos (x – 3/2) +1 = 0 Find its roots on the segment

2. Solve the equation 4cos2 x + 8 cos (x – 3 /2) +1 = 0
Find its roots on the segment
4cos2 x + 8 cos (x – 3 /2) +1 = 0
4cos2x + 8 cos (3 /2 – x) +1 = 0,
4cos2x – 8 sin x +1 = 0,
4 – 4sin2 x – 8 sin x +1 = 0,
4sin 2x + 8sin x – 5 = 0,
D/4 = 16 + 20 = 36,
sin x = – 2.5
or
sin x = ½
x = (-1)k /6 + k, k Z

13. Let’s select roots on a segment (using graphs)

Let's select roots on a segment
(using graphs)
sin x = ½
Let's plot the functions y = sin x and y = ½
x = 4 + /6 = 25 /6
Answer: a) (-1)k /6 + k, k Z; b) 25 /6

14. 3. Solve the equation Find its roots on the segment

4 – cos2 2x = 3 sin2 2x + 2 sin 4x
4 (sin2 2x + cos2 2x) – cos2 2x = 3 sin2 2x + 4 sin 2x cos 2x,
sin2 2x + 3 cos2 2x – 4 sin 2x cos 2x = 0
If cos2 2x = 0, then sin2 2x = 0, which is impossible, so
cos2 2x 0 and both sides of the equation can be divided by cos2 2x.
tg22x + 3 – 4 tg 2x = 0,
tg22x – 4 tg 2x + 3= 0,
tan 2x = 1,
2x = /4 + n, n Z
x = /8 + n/2, n Z
or
tan 2x = 3,
2x = arctan 3 + k, k Z
x = ½ arctan 3 + k/2, k Z

15.

4 – cos2 2x = 3 sin2 2x + 2 sin 4x
x = /8 + n/2, n Z or x = ½ arctan 3 + k/2, k Z
Since 0< arctg 3< /2,
0 < ½ arctg 3< /4, то ½ arctg 3
is the solution
Since 0< /8 < /4 < 1,значит /8
is also a solution
Other solutions will not be included in
gap since they
are obtained from the numbers ½ arctan 3 and /8
adding numbers that are multiples of /2.
Answer: a) /8 + n/2, n Z ; ½ arctan 3 + k/2, k Z
b) /8; ½ arctan 3

16. 4. Solve the equation log5(cos x – sin 2x + 25) = 2 Find its roots on the segment

4. Solve the equation log5(cos x – sin 2x + 25) = 2
Find its roots on the segment
Let's solve the equation:
log5(cos x – sin 2x + 25) = 2
ODZ: cos x – sin 2x + 25 > 0,
cos x – sin 2x + 25 = 25, 25 > 0,
cos x – 2sin x cos x = 0,
cos x (1 – 2sin x) = 0,
cos x = 0,
x = /2 + n, n Z
or
1 – 2sinx = 0,
sin x = 1/2
x = (-1)k /6 + k, k Z

17.

Let's select roots on a segment
Let's select roots on the segment:
1) x = /2 + n, n Z
2 /2 + n 7 /2, n Z
2 1/2 + n 7/2, n Z
2 – ½ n 7/2 – ½, n Z
1.5 n 3, n Z
n = 2; 3
x = /2 + 2 = 5 /2
x = /2 + 3 = 7 /2
2) sin x = 1/2
x = 2 + /6 = 13 /6
x = 3 – /6 = 17 /6
Answer: a) /2 + n, n Z ; (-1)k /6 + k, k Z
b) 13/6; 5 /2; 7 /2; 17/6

18. 5. Solve the equation 1/sin2x + 1/sin x = 2 Find its roots on the segment [-5/2; -3/2]

5. Solve the equation 1/sin2x + 1/sin x = 2
Find its roots on the segment [-5 /2; -3 /2]
Let's solve the equation:
1/sin2x + 1/sin x = 2
x k
Replacement 1/sin x = t,
t2 + t = 2,
t2 + t – 2 = 0,
t1= – 2, t2 = 1
1/sin x = – 2,
sin x = – ½,
x = – /6 + 2 n, n Z
or
x = – 5 /6 + 2 n, n Z
1/sin x = 1,
sin x = 1,
x = /2 + 2 n, n Z
This series of roots is excluded, because -150º+ 360ºn is outside the limits
specified interval [-450º; -270º]

19.

Let's continue selecting roots on the segment
Let's consider the remaining series of roots and carry out a selection of roots
on the segment [-5 /2; -3 /2] ([-450º; -270º]):
1) x = - /6 + 2 n, n Z
2) x = /2 + 2 n, n Z
-5 /2 - /6 + 2 n -3 /2, n Z
-5 /2 /2 + 2 n -3 /2, n Z
-5/2 -1/6 + 2n -3/2, n Z
-5/2 1/2 + 2n -3/2, n Z
-5/2 +1/6 2n -3/2 + 1/6, n Z
-5/2 - 1/2 2n -3/2 - 1/2, n Z
– 7/3 2n -4/3, n Z
– 3 2n -2, n Z
-7/6 n -2/3, n Z
-1.5 n -1. n Z
n = -1
n = -1
x = - /6 - 2 = -13 /6 (-390º)
x = /2 - 2 = -3 /2 (-270º)
Answer: a) /2 + 2 n, n Z ; (-1)k+1 /6 + k, k Z
b) -13 /6; -3 /2

20. 6. Solve the equation |sin x|/sin x + 2 = 2cos x Find its roots on the segment [-1; 8]

Let's solve the equation
|sin x|/sin x + 2 = 2cos x
1)If sin x >0, then |sin x| =sin x
The equation will take the form:
2 cos x=3,
cos x =1.5 – has no roots
2) If sin x<0, то |sin x| =-sin x
and the equation will take the form
2cos x=1, cos x = 1/2,
x = ±π/3 +2πk, k Z
Considering that sin x< 0, то
one series of answers left
x = - π/3 +2πk, k Z
Let's select the roots for
segment [-1; 8]
k=0, x= - π/3 , - π< -3, - π/3 < -1,
-π/3 does not belong to this
segment
k=1, x = - π/3 +2π = 5π/3<8,
5 π/3 [-1; 8]
k=2, x= - π/3 + 4π = 11π/3 > 8,
11π/3 does not belong to this
segment.
Answer: a) - π/3 +2πk, k Z
b) 5
π/3

21. 7. Solve the equation 4sin3x=3cos(x- π/2) Find its roots on the interval

8. Solve the equation √1-sin2x= sin x
Find its roots on the interval
Let's solve the equation √1-sin2x= sin x.
sin x ≥ 0,
1- sin2x = sin2x;
sin x ≥ 0,
2sin2x = 1;
sin x≥0,
sin x =√2/2; sin x = - √2/2;
sin x =√2/2
x=(-1)k /4 + k, k Z
sin x =√2/2

25. Let’s select roots on a segment

Let's select roots on a segment
x=(-1)k /4 + k, k Z
sin x =√2/2
y =sin x and y=√2/2
5 /2 + /4 = 11 /4
Answer: a) (-1)k /4 + k, k Z; b) 11 /4

26. 9. Solve the equation (sin2x + 2 sin2x)/√-cos x =0 Find its roots in the interval [-5; -7/2]

9. Solve the equation (sin2x + 2 sin2x)/√-cos x =0
Find its roots on the interval [-5; -7 /2]
Let's solve the equation
(sin2x + 2 sin2x)/√-cos x =0.
1) ODZ: cos x<0 ,
/2 +2 n 2) sin2x + 2 sin2x =0,
2 sinx∙cos x + 2 sin2x =0,
sin x (cos x+ sin x) =0,
sin x=0, x= n, n Z
or
cos x+ sin x=0 | : cos x,
tan x= -1, x= - /4 + n, n Z
Taking into account DL
x= n, n Z, x= +2 n, n Z;
x= - /4 + n, n Z,
x= 3 /4 + 2 n, n Z

27. Let's select roots on a given segment

Let's select roots on the given
segment [-5; -7 /2]
x= +2 n, n Z ;
-5 ≤ +2 n ≤ -7 /2,
-5-1 ≤ 2n ≤ -7/2-1,
-3≤ n ≤ -9/4, n Z
n = -3, x= -6 = -5
x= 3 /4 + 2 n, n Z
-5 ≤ 3 /4 + 2 n ≤ -7 /2
-23/8 ≤ n ≤ -17/8, no such thing
whole n.
Answer: a) +2 n, n Z ;
3 /4 + 2 n, n Z ;
b) -5.

28. 10. Solve the equation 2sin2x =4cos x –sinx+1 Find its roots on the interval [/2; 3/2]

10. Solve the equation 2sin2x =4cos x –sinx+1
Find its roots on the interval [ /2; 3 /2]
Let's solve the equation
2sin2x = 4cos x – sinx+1
2sin2x = 4cos x – sinx+1,
4 sinx∙cos x – 4cos x + sin x -1 = 0,
4cos x(sin x – 1) + (sin x – 1) = 0,
(sin x – 1)(4cos x +1)=0,
sin x – 1= 0, sin x = 1, x = /2+2 n, n Z
or
4cos x +1= 0, cos x = -0.25
x = ± (-arccos (0.25)) + 2 n, n Z
Let's write the roots of this equation differently
x = - arccos(0.25) + 2 n,
x = -(- arccos(0.25)) + 2 n, n Z

29. Let's select roots using a circle

x = /2+2 n, n Z, x = /2;
x = -arccos(0.25)+2n,
x=-(-arccos(0.25)) +2 n, n Z,
x = - arccos(0.25),
x = + arccos(0.25)
Answer: a) /2+2 n,
-arccos(0.25)+2 n,
-(-arccos(0.25)) +2 n, n Z;
b) /2;
-arccos(0.25); +arccos(0.25)

You can order a detailed solution to your problem!!!

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 an infinite number of solutions.

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 sine, it has no solutions among real numbers.

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 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 using known methods.

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

Solution. Let's write 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`. 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`.

Moving to Half Angle

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

Solution. Let's apply the double angle formulas, 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 algebraic method described above, we obtain:

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 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 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.

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