How to find the roots of the equation belonging to the segment. Finding the roots of the equation belonging to the segment. Methods for solving trigonometric equations

27. Select the roots on a given segment

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

29. Select the roots using a circle

Task #1

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

If we were to solve an equation of the form:

Then we would write the following answer:

Or (because)

But now we are playing the following expression:

Then you can write:

Our goal with you is to make it so that you stand on the left simply, without any "impurities"!

Let's get rid of them!

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

Now we get rid of by dividing both parts by it:

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 it is necessary to sort out.

Let's look at the first series first:

It is clear that if we take, then as a result we will get positive numbers, but we are not interested in them.

So it must be taken negative. Let be.

When the root will be already:

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

Now consider the second series:

And again we substitute: , then:

Not interested!

Then it doesn't make sense to increase it anymore! Let's reduce! Let then:

Fits!

Let be. Then

Then - the largest negative root!

Answer:

Task #2

Again, we solve, regardless of the complex cosine argument:

Now we express again on the left:

Multiply both sides by

Divide both sides

All that's left 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. Consider the first series:

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

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 #3

We decide, regardless of the complex argument of the tangent.

That seems to be nothing complicated, right?

As before, we express on the left side:

Well, that's great, there is generally only one series of roots! Again, find the largest negative.

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

Answer:

Now try to solve the following problems on your own.

Homework or 3 tasks for independent solution.

1. Re-shi-te equation.
   
2. Re-shi-te equation.
   In from-ve-te on-pi-shi-te the smallest in-lo-zhi-tel-ny root.
3. Re-shi-te equation.
   In from-ve-te on-pi-shi-te the smallest in-lo-zhi-tel-ny root.

Ready? We check. I will not describe in detail the entire solution algorithm, it seems to me that enough attention has already been paid to it above.

Well, is everything right? Oh, those nasty sinuses, there are always some troubles with them!

Well, now you can solve the simplest trigonometric equations!

Check out the solutions and answers:

Task #1

Express

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

Answer:

Task #2

The smallest positive root will be obtained at.

He will be equal.

Answer: .

Task #3

When we get, when we have.

Answer: .

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

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

AVERAGE LEVEL

In this article I will describe solution of trigonometric equations of a more complex type and how to select their roots. Here I will focus on the following topics:

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

More complex trigonometric equations are the basis of problems of increased complexity. They require both solving the equation itself in general form and finding the roots of this equation that belong to some given interval.

The solution of trigonometric equations is reduced to two subtasks:

1. Equation solution
2. Root selection

It should be noted that the second is not always required, but still in most examples it is required to make a selection. And if it is not required, then you can rather sympathize - this means that the equation is quite complicated in itself.

My experience with the analysis of C1 tasks 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 that reduce to the form.
3. Equations Solved by Change of Variable.
4. Equations requiring additional selection of roots due to irrationality or denominator.

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

If you come across an equation of type 4, then you are less fortunate: 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 I will devote this one to solving equations of the first three types.

Equations Reducing to Factoring

The most important thing you need to remember in order to solve equations of this type is

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

Example 1. An equation that reduces to factorization using the formulas of reduction and the sine of a double angle

• Re-shi-te equation
• Find-di-those all the roots of this equation

Here, as I promised, the casting formulas work:

Then my equation will look like this:

Then my equation will take the following form:

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

So what to do? Yes, everything is simple, transfer everything in one direction and take out the common factor:

Well, we factored it out, hooray! Now we decide:

The first equation has roots:

And the second:

This completes the first part of the problem. Now we 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 series (and it's easier, 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 - a bit too much, it doesn't fit.

Let, then - again did not hit.

One more try - then - there, hit! First root found!

I shoot again: then - hit again!

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

So from the first series, 2 roots belong to the interval: .

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

Undershoot!

Missing again!

Again shortfall!

Got it!

Flight!

Thus, the following roots belong to my span:

We will use this algorithm to solve all other examples. Let's practice one more example together.

Example 2. An equation that reduces to factorization using reduction formulas

• Solve the Equation

Solution:

Again the notorious cast formulas:

Again, do not try to cut!

The first equation has roots:

And the second:

Now again the search for roots.

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

Now the first series and it is simpler:

If - suitable

If - also good

If - already flight.

Then the roots will be:

Independent work. 3 equations.

Well, do you understand the technique? Solving trigonometric equations no longer seems so difficult? Then quickly solve the following problems yourself, and then you and I will solve other examples:

1. Solve the Equation
   Find all the roots of this equation that are attached to the gap.
2. Re-shi-te equation
   Indicate the roots of the equation, which are attached to the cut
3. Re-shi-te equation
   Find-di-those all the roots of this equation, at-above-le-zha-shchi pro-inter-zhut-ku.

Equation 1

And again the casting formula:

First series of roots:

Second series of roots:

We start the selection for the interval

Answer: , .

Equation 2 Checking independent work.

Pretty tricky grouping into factors (I'll use the formula for the sine of a double angle):

then or

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

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

Let's make 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. Verification of independent work.

A frightening equation. However, it is solved quite simply by applying the formula for the sine of a double angle:

Let's cut it down by 2:

We 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 consider the second:

In general, I was going to dwell on solving such equations a little later, but since it turned up, there was nothing to do, we had to decide ...

Equations of the form:

This equation is solved by dividing both sides by:

Thus, our equation has a single series of roots:

You need to find those of them that belong to the interval: .

Let's build the table again, as I did before:

Answer: .

Equations that reduce to the form:

Well, now it's time to move on to the second portion of the equations, especially since I already blurted out what the solution of the new type of trigonometric equations consists of. But it will not be superfluous to repeat that the equation of the form

It is solved by dividing both parts by the cosine:

1. Re-shi-te equation
   Indicate the roots of the equation that are attached to the cut-off.
2. Re-shi-te equation
   Indicate the roots of the equation, at-above-le-zha-shchi pro-inter-zhut-ku.

Example 1

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

Aha! Type equation: . I divide both parts into

We do root elimination:

Gap:

Answer:

Example 2

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

Basic trigonometric identity:

Sine of a double angle:

Finally we get:

Screening of roots: gap.

Answer: .

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

• Re-shi-te equation
• Find-di-those all the roots of this equation, at-above-le-zha-schie from-cut.

Let's check:

The equation is solved immediately, it is enough to divide both parts by:

Root sifting:

Answer: .

One way or another, we have yet to encounter equations of the kind we have just discussed. However, it is still too early for us to wrap up: there is one more "layer" of equations that we have not analyzed. So:

Solution of trigonometric equations by change of variable

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

Example.

• Solve the equation: .
• Find-di-those all the roots of this equation, at-above-le-zha-schie from-cut.

Well, here the replacement itself suggests itself into our hands!

Then our equation becomes this:

The first equation has roots:

And the second one is like this:

Now let's find the roots that belong to the interval

Answer: .

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

• Re-shi-te equation
• Indicate the roots of the given equation, at-above-le-zha-shchi pro-inter-zhut-ku.

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

We can, for example, imagine

And at the same time

Then my equation becomes:

And now attention, focus:

Let's divide both sides of the equation into:

Suddenly, you and I got a quadratic equation for! Let's make a substitution, then we get:

The equation has the following roots:

An unpleasant second series of roots, but there's nothing to be done! We make a selection of roots on the interval.

We also need to take into account that

Since and then

Answer:

To consolidate, before you solve the problems yourself, here's another exercise for you:

• Re-shi-te equation
• Find-di-those all the roots of this equation, at-above-le-zha-shchi pro-inter-zhut-ku.

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

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

Now I will go from cosine to sine according to the basic trigonometric identity:

And finally, I will bring everything to a common denominator:

Now I can go to the equation:

But at (i.e. at).

Now everything is ready for replacement:

Then either

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

Who suffers from this? The trouble is with the tangent, it is not defined when the cosine is zero (division by zero occurs).

So the roots of the equation are:

Now we screen out the roots in the interval:

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

You see: the appearance of the denominator (as well as the tangent, leads to certain difficulties with the roots! You need to be more careful here!).

Well, you and I have almost finished the analysis of trigonometric equations, there is very little left - to solve two problems on our own. Here they are.

1. Solve the Equation
   Find-di-those all the roots of this equation, at-above-le-zha-schie from-cut.
2. Re-shi-te equation
   Indicate the roots of this equation, which are attached to the cut.

Decided? Not very difficult? Let's check:

1. We work according to the reduction formulas:

   We substitute into the equation:

   Let's rewrite everything in terms of cosines, so that it is more convenient to make the replacement:

   Now it's easy to make the substitution:

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

   We are looking for the roots we need on the interval

   Answer: .

2. 
   Here the replacement is immediately visible:

   Then either

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

   Answer:

Well, now everything! But the solution of trigonometric equations does not end there, we left behind the most difficult cases: when there is irrationality or various kinds of “complex denominators” in the equations. How to solve such tasks, we will consider in an article for an advanced level.

ADVANCED LEVEL

In addition to the trigonometric equations considered in the previous two articles, we consider another class of equations that require even more careful analysis. These trigonometric examples contain either an 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, there is a silver lining: for such equations, as a rule, the question of which of its roots belong to a given interval is no longer raised. Let's not beat around the bush, but just trigonometric examples.

Example 1

Solve the equation and find those 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:

Now let's look at the series:

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

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

Now we select the roots belonging to the interval.

Then the roots are:

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

Example 2

Solve the equation:

Solution:

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

And what, is that all? No, alas, that would be too easy! It must be remembered that only non-negative numbers can stand under the root. Then:

Solution to this inequality:

Now it remains to find out if a part of the roots of the first equation did not inadvertently fall into a place where the inequality does not hold.

To do this, you can again use the table:

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

Answer:

You see, the root requires even closer attention! Let's complicate: let now I have a trigonometric function under the root.

Example 3

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

Now the second equation:

Now the most difficult thing is to find out if negative values ​​\u200b\u200bare 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 about degrees, radians are about 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 about the expression:

It's less than zero!

So - is not the root of the equation.

Now turn.

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 about degrees. In the same time

since, then, and therefore
,

Answer: .

Could it be even more difficult? 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, look further:

Example 4

The root is not suitable, due to the limited cosine

Now the second one:

At the same time, by definition of the root:

We must 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 quadrant.

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

Answer: ,

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

Example 5

Well, there's nothing to be done - we act as before.

Now we work with the denominator:

I don’t want to solve the trigonometric inequality, and therefore I’ll do it tricky: I’ll take and substitute my series of roots into the inequality:

If is even, then we have:

since, then all the 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 is odd, then:

What quarter is the angle in? This is the corner of the second quarter. Then all the corners are again the corners of the second quarter. The sine is positive. Just what you need! So the series is:

Fits!

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

Substitute into our inequality:

If is even, then

Corners of the first quarter. The sine is positive there, so the series is suitable. Now if is odd, then:

fits too!

Well, now we write down the answer!

Answer:

Well, this was perhaps the most laborious case. Now I offer you tasks for independent solution.

Training

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

Solutions:

1. 
   First equation:
   or
   Root ODZ:

   Second equation:

   Selection of roots that belong to the interval

   Answer:

Or
or
But

Consider: . If is even, then
- does not fit!
If - odd, : - fits!
So our equation has the following series of roots:
or
Selection of roots on the interval:

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

Answer: , .

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

Second part:

At the same time, ODZ requires that

We check the roots found in the first equation:

If sign:

Angles of the first quarter, where the tangent is positive. Not suitable!
If sign:

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

Answer: , .

We've broken down complex trigonometric examples together in this article, but you should be able to solve the equations yourself.

SUMMARY AND BASIC FORMULA

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 a trigonometric circle.

Allows you to measure angles, find their sines, cosines, and more.

A) Solve the equation 2(\sin x-\cos x)=tgx-1.

b) \left[ \frac(3\pi )2;\,3\pi \right].

Solution

A) Opening the brackets and moving all the terms to the left side, we get the equation 1+2 \sin x-2 \cos x-tg x=0. Considering that \cos x \neq 0, the term 2 \sin x can be replaced by 2 tg x \cos x, we obtain the equation 1+2 tan x \cos x-2 \cos x-tg x=0, which, by grouping, can be reduced to the form (1-tg x)(1-2 \cos x)=0.

1) 1-tgx=0, tanx=1, x=\frac\pi 4+\pi n, n \in \mathbb Z;

2) 1-2 \cos x=0, \cosx=\frac12, x=\pm \frac\pi 3+2\pi n, n \in \mathbb Z.

b) With the help of a numerical circle, we select the roots belonging to the interval \left[ \frac(3\pi )2;\, 3\pi \right].

x_1=\frac\pi 4+2\pi =\frac(9\pi )4,

x_2=\frac\pi 3+2\pi =\frac(7\pi )3,

x_3=-\frac\pi 3+2\pi =\frac(5\pi )3.

Answer

A) \frac\pi 4+\pi n, \pm\frac\pi 3+2\pi n, n \in \mathbb Z;

b) \frac(5\pi )3, \frac(7\pi )3, \frac(9\pi )4.

Condition

A) Solve the Equation (2\sin ^24x-3\cos 4x)\cdot \sqrt (tgx)=0.

b) Indicate the roots of this equation that belong to the interval \left(0;\,\frac(3\pi )2\right] ;

Solution

A) ODZ: \begin(cases) tgx\geqslant 0\\x\neq \frac\pi 2+\pi k,k \in \mathbb Z. \end(cases)

The original equation on the ODZ is equivalent to the set of equations

\left[\!\!\begin(array)(l) 2 \sin ^2 4x-3 \cos 4x=0,\\tg x=0. \end(array)\right.

Let's solve the first equation. To do this, we will replace \cos 4x=t, t \in [-1; 1]. Then \sin^24x=1-t^2. We get:

2(1-t^2)-3t=0,

2t^2+3t-2=0,

t_1=\frac12, t_2=-2, t_2\notin [-1; 1].

\cos4x=\frac12,

4x=\pm \frac\pi 3+2\pi n,

x=\pm \frac\pi (12)+\frac(\pi n)2, n \in \mathbb Z.

Let's solve the second equation.

tg x=0,\, x=\pi k, k \in \mathbb Z.

Using the unit circle, we find solutions that satisfy the ODZ.

The sign "+" marks the 1st and 3rd quarters, in which tg x>0.

We get: x=\pi k, k \in \mathbb Z; x=\frac\pi (12)+\pi n, n \in \mathbb Z; x=\frac(5\pi )(12)+\pi m, m \in \mathbb Z.

b) Let's find the roots belonging to the interval \left(0;\,\frac(3\pi )2\right].

x=\frac\pi (12), x=\frac(5\pi )(12); x=\pi ; x=\frac(13\pi )(12); x=\frac(17\pi )(12).

Answer

A) \pi k, k \in \mathbb Z; \frac\pi (12)+\pi n, n \in \mathbb Z; \frac(5\pi )(12)+\pi m, m \in \mathbb Z.

b) \pi; \frac\pi(12); \frac(5\pi )(12); \frac(13\pi )(12); \frac(17\pi )(12).

Source: "Mathematics. Preparation for the exam-2017. profile level. Ed. F. F. Lysenko, S. Yu. Kulabukhova.

Condition

A) Solve the equation: \cos ^2x+\cos ^2\frac\pi 6=\cos ^22x+\sin ^2\frac\pi 3;

b) Specify all roots belonging to the interval \left(\frac(7\pi )2;\,\frac(9\pi )2\right].

Solution

A) Because \sin \frac\pi 3=\cos \frac\pi 6, That \sin ^2\frac\pi 3=\cos ^2\frac\pi 6, hence, the given equation is equivalent to the equation \cos^2x=\cos ^22x, which, in turn, is equivalent to the equation \cos^2x-\cos ^2 2x=0.

But \cos ^2x-\cos ^22x= (\cos x-\cos 2x)\cdot (\cos x+\cos 2x) And

\cos 2x=2 \cos ^2 x-1, so the equation becomes

(\cos x-(2 \cos ^2 x-1))\,\cdot(\cos x+(2 \cos ^2 x-1))=0,

(2 \cos ^2 x-\cos x-1)\,\cdot (2 \cos ^2 x+\cos x-1)=0.

Then either 2 \cos ^2 x-\cos x-1=0 or 2 \cos ^2 x+\cos x-1=0.

Solving the first equation as a quadratic equation for \cos x, we get:

(\cos x)_(1,2)=\frac(1\pm\sqrt 9)4=\frac(1\pm3)4. Therefore, either \cos x=1 or \cosx=-\frac12. If \cos x=1, then x=2k\pi , k \in \mathbb Z. If \cosx=-\frac12, That x=\pm \frac(2\pi )3+2s\pi , s \in \mathbb Z.

Similarly, solving the second equation, we get either \cos x=-1, or \cosx=\frac12. If \cos x=-1, then the roots x=\pi +2m\pi , m \in \mathbb Z. If \cosx=\frac12, That x=\pm \frac\pi 3+2n\pi , n \in \mathbb Z.

Let's combine the obtained solutions:

x=m\pi , m \in \mathbb Z; x=\pm \frac\pi 3 +s\pi , s \in \mathbb Z.

b) We select the roots that fall within the given interval using a number circle.

We get: x_1 =\frac(11\pi )3, x_2=4\pi , x_3 =\frac(13\pi )3.

Answer

A) m\pi, m \in \mathbb Z; \pm \frac\pi 3 +s\pi , s \in \mathbb Z;

b) \frac(11\pi )3, 4\pi , \frac(13\pi )3.

Source: "Mathematics. Preparation for the exam-2017. profile level. Ed. F. F. Lysenko, S. Yu. Kulabukhova.

Condition

A) Solve the Equation 10\cos ^2\frac x2=\frac(11+5ctg\left(\dfrac(3\pi )2-x\right) )(1+tgx).

b) Indicate the roots of this equation that belong to the interval \left(-2\pi ; -\frac(3\pi )2\right).

Solution

A) 1. According to the reduction formula, ctg\left(\frac(3\pi )2-x\right) =tgx. The domain of the equation will be x values ​​such that \cos x \neq 0 and tg x \neq -1. We transform the equation using the double angle cosine formula 2 \cos ^2 \frac x2=1+\cos x. We get the equation: 5(1+\cos x) =\frac(11+5tgx)(1+tgx).

notice, that \frac(11+5tgx)(1+tgx)= \frac(5(1+tgx)+6)(1+tgx)= 5+\frac(6)(1+tgx), so the equation becomes: 5+5 \cos x=5 +\frac(6)(1+tgx). From here \cosx=\frac(\dfrac65)(1+tgx), \cosx+\sinx=\frac65.

2. Transform \sin x+\cos x using the reduction formula and the formula for the sum of cosines: \sin x=\cos \left(\frac\pi 2-x\right), \cos x+\sin x= \cos x+\cos \left(\frac\pi 2-x\right)= 2\cos \frac\pi 4\cos \left(x-\frac\pi 4\right)= \sqrt 2\cos \left(x-\frac\pi 4\right) = \frac65.

From here \cos \left(x-\frac\pi 4\right) =\frac(3\sqrt 2)5. Means, x-\frac\pi 4= arc\cos \frac(3\sqrt 2)5+2\pi k, k \in \mathbb Z,

or x-\frac\pi 4= -arc\cos \frac(3\sqrt 2)5+2\pi t, t \in \mathbb Z.

That's why x=\frac\pi 4+arc\cos \frac(3\sqrt 2)5+2\pi k,k \in \mathbb Z,

or x =\frac\pi 4-arc\cos \frac(3\sqrt 2)5+2\pi t,t \in \mathbb Z.

The found values ​​of x belong to the domain of definition.

b) Let us first find out where the roots of the equation fall at k=0 and t=0. These will be respectively the numbers a=\frac\pi 4+arccos \frac(3\sqrt 2)5 And b=\frac\pi 4-arccos \frac(3\sqrt 2)5.

1. Let us prove an auxiliary inequality:

\frac(\sqrt 2)(2)<\frac{3\sqrt 2}2<1.

Really, \frac(\sqrt 2)(2)=\frac(5\sqrt 2)(10)<\frac{6\sqrt2}{10}=\frac{3\sqrt2}{5}.

Note also that \left(\frac(3\sqrt 2)5\right) ^2=\frac(18)(25)<1^2=1, Means \frac(3\sqrt 2)5<1.

2. From inequalities (1) by the property of the arccosine we get:

arccos 1

0

From here \frac\pi 4+0<\frac\pi 4+arc\cos \frac{3\sqrt 2}5<\frac\pi 4+\frac\pi 4,

0<\frac\pi 4+arccos \frac{3\sqrt 2}5<\frac\pi 2,

0

Likewise, -\frac\pi 4

0=\frac\pi 4-\frac\pi 4<\frac\pi 4-arccos \frac{3\sqrt 2}5< \frac\pi 4<\frac\pi 2,

0

With k=-1 and t=-1 we get the roots of the equation a-2\pi and b-2\pi.

\Bigg(a-2\pi =-\frac74\pi +arccos \frac(3\sqrt 2)5,\, b-2\pi =-\frac74\pi -arccos \frac(3\sqrt 2)5\Bigg). Wherein -2\pi

2\pi So these roots belong to the given interval \left(-2\pi , -\frac(3\pi )2\right).

For other values ​​of k and t, the roots of the equation do not belong to the given interval.

Indeed, if k\geqslant 1 and t\geqslant 1, then the roots are greater than 2\pi. If k\leqslant -2 and t\leqslant -2, then the roots are less -\frac(7\pi )2.

Answer

A) \frac\pi4\pm arccos\frac(3\sqrt2)5+2\pi k, k\in\mathbb Z;

b) -\frac(7\pi)4\pm arccos\frac(3\sqrt2)5.

Source: "Mathematics. Preparation for the exam-2017. profile level. Ed. F. F. Lysenko, S. Yu. Kulabukhova.

Condition

A) Solve the Equation \sin \left(\frac\pi 2+x\right) =\sin (-2x).

b) Find all the roots of this equation that belong to the interval ;

Solution

A) Let's transform the equation:

\cosx=-\sin 2x,

\cos x+2 \sin x \cos x=0,

\cos x(1+2\sin x)=0,

\cosx=0,

x =\frac\pi 2+\pi n, n \in \mathbb Z;

1+2\sinx=0,

\sinx=-\frac12,

x=(-1)^(k+1)\cdot \frac\pi 6+\pi k, k \in \mathbb Z.

b) We find the roots belonging to the segment using the unit circle.

The specified interval contains a single number \frac\pi 2.

Answer

A) \frac\pi 2+\pi n, n \in \mathbb Z; (-1)^(k+1)\cdot \frac\pi 6+\pi k, k \in \mathbb Z;

b) \frac\pi 2.

Source: "Mathematics. Preparation for the exam-2017. profile level. Ed. F. F. Lysenko, S. Yu. Kulabukhova.

Condition

Means, \sin x \neq 1.

Divide both sides of the equation by the factor (\sinx-1), different from zero. We get the equation \frac 1(1+\cos 2x)=\frac 1(1+\cos (\pi +x)), or equation 1+\cos 2x=1+\cos (\pi +x). Applying the reduction formula on the left side, and the reduction formula on the right side, we obtain the equation 2 \cos ^2 x=1-\cos x. This is the equation using the substitution \cosx=t, Where -1 \leqslant t \leqslant 1 reduce to square: 2t^2+t-1=0, whose roots t_1=-1 And t_2=\frac12. Returning to the variable x , we get \cos x = \frac12 or \cosx=-1, where x=\frac \pi 3+2\pi m, m \in \mathbb Z, x=-\frac \pi 3+2\pi n, n \in \mathbb Z, x=\pi +2\pi k, k \in \mathbb Z.

b) Solve inequalities

1) -\frac(3\pi )2 \leqslant \frac(\pi )3+2\pi m \leqslant -\frac \pi 2 ,

2) -\frac(3\pi )2 \leqslant -\frac \pi 3+2\pi n \leqslant -\frac \pi (2,)

3) -\frac(3\pi )2 \leqslant \pi+2\pi k \leqslant -\frac \pi 2 , m, n, k \in \mathbb Z.

1) -\frac(3\pi )2 \leqslant \frac(\pi )3+2\pi m \leqslant -\frac \pi 2 , -\frac32\leqslant \frac13+2m \leqslant -\frac12 -\frac(11)6 \leqslant 2m \leqslant -\frac56 , -\frac(11)(12) \leqslant m \leqslant -\frac5(12).

\left [-\frac(11)(12);-\frac5(12)\right].

2) -\frac (3\pi) 2 \leqslant -\frac(\pi )3+2\pi n \leqslant -\frac(\pi )(2), -\frac32 \leqslant -\frac13 +2n \leqslant -\frac12 , -\frac76 \leqslant 2n \leqslant -\frac1(6), -\frac7(12) \leqslant n \leqslant -\frac1(12).

There are no integers belonging to the interval \left[-\frac7(12) ; -\frac1(12)\right].

3) -\frac(3\pi )2 \leqslant \pi +2\pi k\leqslant -\frac(\pi )2, -\frac32 \leqslant 1+2k\leqslant -\frac12, -\frac52 \leqslant 2k \leqslant -\frac32, -\frac54 \leqslant k \leqslant -\frac34.

This inequality is satisfied by k=-1, then x=-\pi.

Answer

A) \frac \pi 3+2\pi m; -\frac \pi 3+2\pi n; \pi +2\pi k, m, n, k \in \mathbb Z;

b) -\pi .

In this article I will try to explain 2 ways taking roots in a trigonometric equation: using inequalities and using a trigonometric circle. Let's move on to a clear example and we'll figure it out as we go.

A) Solve the equation sqrt(2)cos^2x=sin(Pi/2+x)
b) Find all the roots of this equation that belong to the interval [-7Pi/2; -2Pi]

Let's solve a.

We use the reduction formula for the sine sin(Pi/2+x) = cos(x)

Sqrt(2)cos^2x = cosx

Sqrt(2)cos^2x - cosx = 0

Cosx(sqrt(2)cosx - 1) = 0

X1 = Pi/2 + Pin, n ∈ Z

Sqrt(2)cos - 1 = 0

cox = 1/sqrt(2)

Cox = sqrt(2)/2

X2 = arccos(sqrt(2)/2) + 2Pin, n ∈ Z
x3 = -arccos(sqrt(2)/2) + 2Pin, n ∈ Z

X2 = Pi/4 + 2Pin, n ∈ Z
x3 = -Pi/4 + 2Pin, n ∈ Z

Let's solve point b.

1) Selection of roots using inequalities

Here everything is done simply, we substitute the obtained roots into the interval given to us [-7Pi / 2; -2Pi], find integer values ​​for n.

7Pi/2 is less than or equal to Pi/2 + Pin is less than or equal to -2Pi

Immediately divide everything by Pi

7/2 less than or equal to 1/2 + n less than or equal to -2

7/2 - 1/2 less than or equal to n less than or equal to -2 - 1/2

4 less than or equal to n less than or equal to -5/2

The integers n in this gap are -4 and -3. So the roots belonging to this interval will be Pi/2 + Pi(-4) = -7Pi/2, Pi/2 + Pi(-3) = -5Pi/2

Similarly, we make two more inequalities

7Pi/2 is less than or equal to Pi/4 + 2Pin is less than or equal to -2Pi
-15/8 less than or equal to n less than or equal to -9/8

There are no integers n in this interval

7Pi/2 less than or equal to -Pi/4 + 2Pin less than or equal to -2Pi
-13/8 less than or equal to n less than or equal to -7/8

One integer n in this gap is -1. So the selected root on this interval is -Pi/4 + 2Pi*(-1) = -9Pi/4.

So the answer in paragraph b: -7Pi / 2, -5Pi / 2, -9Pi / 4

2) Selection of roots using a trigonometric circle

To use this method, you need to understand how this circle works. I'll try to explain in simple terms how I understand it. I think in schools in algebra lessons this topic was explained many times by the clever words of the teacher, in textbooks there are complex formulations. Personally, I understand this as a circle that can be walked around an infinite number of times, this is explained by the fact that the sine and cosine functions are periodic.

Let's go around counterclockwise

Go around 2 times counterclockwise

Go around 1 time clockwise (values ​​will be negative)

Let's return to our question, we need to select the roots on the interval [-7Pi/2; -2Pi]

To get to the numbers -7Pi / 2 and -2Pi, you need to go around the circle counterclockwise twice. In order to find the roots of the equation on this interval, it is necessary to estimate and substitute.

Consider x = Pi/2 + Pin. What is the approximate value of n for x to be somewhere in that range? We substitute, let's say -2, we get Pi / 2 - 2Pi = -3Pi / 2, obviously this is not included in our range, so we take less than -3, Pi / 2 - 3Pi = -5Pi / 2, this is suitable, let's try another -4 , Pi/2 - 4Pi = -7Pi/2, also suitable.

Arguing similarly for Pi/4 + 2Pin and -Pi/4 + 2Pin, we find another root -9Pi/4.

Comparison of two methods.

The first method (using inequalities) is much more reliable and much easier to understand, but if you really seriously understand the trigonometric circle and the second selection method, then root selection will be much faster, you can save about 15 minutes on the exam.