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!! So go to negative side 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.
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!
Let's express
The smallest positive root is obtained if we put, since, then
Answer:
The smallest positive root is obtained at.
It will be equal.
Answer: .
When we get, 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).
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:
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 a certain given interval.
Solving trigonometric equations comes down to two subtasks:
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.
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.
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:
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 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.
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:
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:
And again the reduction formula:
First series of roots:
Second series of roots:
We begin selection for the gap
Answer: , .
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 can't say exact value 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
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 it’s time to move on to the second portion of equations, especially since I’ve 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:
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 pure form equations that immediately reduce to an equation for tangent are quite rare. Typically, this transition (division by cosine) is only part of more difficult task. Here's an example for you to practice:
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:
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.
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:
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 got quadratic equation relatively! 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:
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.
Decided? Isn't it very difficult? Let's check:
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: .
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 with the most complex cases: when there is irrationality or various kinds of “complicated denominators” in the equations. We will look at how to solve such tasks in an article for an 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 it now stand under my root trigonometric function.
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 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’s 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 clever: 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.
Solutions:
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.
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.
Quite often in problems of increased complexity we encounter trigonometric equations containing modulus. Most of them require a heuristic approach to solution, which is completely unfamiliar to most schoolchildren.
The problems proposed below are intended to introduce you to the most typical techniques for solving trigonometric equations containing a modulus.
Problem 1. Find the difference (in degrees) of the smallest positive and largest negative roots of the equation 1 + 2sin x |cos x| = 0.
Solution.
Let's expand the module:
1) If cos x ≥ 0, then the original equation will take the form 1 + 2sin x cos x = 0.
Using the double angle sine formula, we get:
1 + sin 2x = 0; sin 2x = -1;
2x = -π/2 + 2πn, n € Z;
x = -π/4 + πn, n € Z. Since cos x ≥ 0, then x = -π/4 + 2πk, k € Z.
2) If cos x< 0, то заданное уравнение имеет вид 1 – 2sin x · cos x = 0. По формуле синуса двойного угла, имеем:
1 – sin 2x = 0; sin 2x = 1;
2x = π/2 + 2πn, n € Z;
x = π/4 + πn, n € Z. Since cos x< 0, то x = 5π/4 + 2πk, k € Z.
3) The largest negative root of the equation: -π/4; smallest positive root of the equation: 5π/4.
The required difference: 5π/4 – (-π/4) = 6π/4 = 3π/2 = 3 180°/2 = 270°.
Answer: 270°.
Problem 2. Find (in degrees) the smallest positive root of the equation |tg x| + 1/cos x = tan x.
Solution.
Let's expand the module:
1) If tan x ≥ 0, then
tan x + 1/cos x = tan x;
The resulting equation has no roots.
2) If tg x< 0, тогда
Tg x + 1/cos x = tg x;
1/cos x – 2tg x = 0;
1/cos x – 2sin x / cos x = 0;
(1 – 2sin x) / cos x = 0;
1 – 2sin x = 0 and cos x ≠ 0.
Using Figure 1 and the condition tg x< 0 находим, что x = 5π/6 + 2πn, где n € Z.
3) The smallest positive root of the equation is 5π/6. Let's convert this value to degrees:
5π/6 = 5 180°/6 = 5 30° = 150°.
Answer: 150°.
Problem 3. Find the number of different roots of the equation sin |2x| = cos 2x on the interval [-π/2; π/2].
Solution.
Let's write the equation in the form sin|2x| – cos 2x = 0 and consider the function y = sin |2x| – cos 2x. Since the function is even, we will find its zeros for x ≥ 0.
sin 2x – cos 2x = 0; Let's divide both sides of the equation by cos 2x ≠ 0, we get:
tg 2x – 1 = 0;
2x = π/4 + πn, n € Z;
x = π/8 + πn/2, n € Z.
Using the parity of the function, we find that the roots of the original equation are numbers of the form
± (π/8 + πn/2), where n € Z.
Interval [-π/2; π/2] belong to the numbers: -π/8; π/8.
So, two roots of the equation belong to the given interval.
Answer: 2.
This equation could also be solved by opening the module.
Problem 4. Find the number of roots of the equation sin x – (|2cos x – 1|)/(2cos x – 1) · sin 2 x = sin 2 x on the interval [-π; 2π].
Solution.
1) Consider the case when 2cos x – 1 > 0, i.e. cos x > 1/2, then the equation takes the form:
sin x – sin 2 x = sin 2 x;
sin x – 2sin 2 x = 0;
sin x(1 – 2sin x) = 0;
sin x = 0 or 1 – 2sin x = 0;
sin x = 0 or sin x = 1/2.
Using Figure 2 and the condition cos x > 1/2, we find the roots of the equation:
x = π/6 + 2πn or x = 2πn, n € Z.
2) Consider the case when 2cos x – 1< 0, т.е. cos x < 1/2, тогда исходное уравнение принимает вид:
sin x + sin 2 x = sin 2 x;
x = 2πn, n € Z.
Using Figure 2 and the cos x condition< 1/2, находим, что x = π + 2πn, где n € Z.
Combining the two cases, we get:
x = π/6 + 2πn or x = πn.
3) Interval [-π; 2π] belong to the roots: π/6; -π; 0; π; 2π.
Thus, the given interval contains five roots of the equation.
Answer: 5.
Problem 5. Find the number of roots of the equation (x – 0.7) 2 |sin x| + sin x = 0 on the interval [-π; 2π].
Solution.
1) If sin x ≥ 0, then the original equation takes the form (x – 0.7) 2 sin x + sin x = 0. After taking the common factor sin x out of brackets, we get:
sin x((x – 0.7) 2 + 1) = 0; since (x – 0.7) 2 + 1 > 0 for all real x, then sinx = 0, i.e. x = πn, n € Z.
2) If sin x< 0, то -(x – 0,7) 2 sin x + sin x = 0;
sin x((x – 0.7) 2 – 1) = 0;
sinx = 0 or (x – 0.7) 2 + 1 = 0. Since sin x< 0, то (x – 0,7) 2 = 1. Извлекаем Square root from the left and right sides of the last equation, we get:
x – 0.7 = 1 or x – 0.7 = -1, which means x = 1.7 or x = -0.3.
Taking into account the condition sinx< 0 получим, что sin (-0,3) ≈ sin (-17,1°) < 0 и sin (1,7) ≈ sin (96,9°) >0, which means only the number -0.3 is the root of the original equation.
3) Interval [-π; 2π] belong to the numbers: -π; 0; π; 2π; -0.3.
Thus, the equation has five roots on a given interval.
Answer: 5.
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