The simplest trigonometric equations are usually solved by formulas. Let me remind you that the following trigonometric equations are called the simplest:
sinx = a
cosx = a
tgx = a
ctgx = a
x is the angle to be found,
a is any number.
And here are the formulas with which you can immediately write down the solutions of these simplest equations.
For sinus:
For cosine:
x = ± arccos a + 2π n, n ∈ Z
For tangent:
x = arctg a + π n, n ∈ Z
For cotangent:
x = arcctg a + π n, n ∈ Z
Actually, this is theoretical part solutions of the simplest trigonometric equations. And, the whole!) Nothing at all. However, the number of errors on this topic just rolls over. Especially, with a slight deviation of the example from the template. Why?
Yes, because a lot of people write down these letters, without understanding their meaning at all! With apprehension he writes down, no matter how something happens ...) This needs to be sorted out. Trigonometry for people, or people for trigonometry, after all!?)
Let's figure it out?
One angle will be equal to arccos a, second: -arccos a.
And that's how it will always work. For any a.
If you don't believe me, hover your mouse over the picture, or touch the picture on your tablet.) I changed the number a to some negative. Anyway, we got one corner arccos a, second: -arccos a.
Therefore, the answer can always be written as two series of roots:
x 1 = arccos a + 2π n, n ∈ Z
x 2 = - arccos a + 2π n, n ∈ Z
We combine these two series into one:
x= ± arccos a + 2π n, n ∈ Z
And all things. We have obtained a general formula for solving the simplest trigonometric equation with cosine.
If you understand that this is not some kind of super-scientific wisdom, but just an abbreviated record of two series of answers, you and tasks "C" will be on the shoulder. With inequalities, with the selection of roots from a given interval ... There, the answer with plus / minus does not roll. And if you treat the answer businesslike, and break it into two separate answers, everything is decided.) Actually, for this we understand. What, how and where.
In the simplest trigonometric equation
sinx = a
also get two series of roots. Always. And these two series can also be recorded one line. Only this line will be smarter:
x = (-1) n arcsin a + π n, n ∈ Z
But the essence remains the same. Mathematicians simply constructed a formula to make one instead of two records of series of roots. And that's it!
Let's check the mathematicians? And that's not enough...)
In the previous lesson, the solution (without any formulas) of the trigonometric equation with a sine was analyzed in detail:
The answer turned out to be two series of roots:
x 1 = π /6 + 2π n, n ∈ Z
x 2 = 5π /6 + 2π n, n ∈ Z
If we solve the same equation using the formula, we get the answer:
x = (-1) n arcsin 0.5 + π n, n ∈ Z
Actually, this is a half-finished answer.) The student must know that arcsin 0.5 = π /6. The full answer would be:
x = (-1) n π /6+ πn, n ∈ Z
Here an interesting question arises. Reply via x 1; x 2 (this is the correct answer!) and through the lonely X (and this is the correct answer!) - the same thing, or not? Let's find out now.)
Substitute in response with x 1 values n =0; one; 2; etc., we consider, we get a series of roots:
x 1 \u003d π / 6; 13π/6; 25π/6 etc.
With the same substitution in response to x 2 , we get:
x 2 \u003d 5π / 6; 17π/6; 29π/6 etc.
And now we substitute the values n (0; 1; 2; 3; 4...) into the general formula for the lonely X . That is, we raise minus one to the zero power, then to the first, second, and so on. And, of course, we substitute 0 into the second term; one; 2 3; 4 etc. And we think. We get a series:
x = π/6; 5π/6; 13π/6; 17π/6; 25π/6 etc.
That's all you can see.) The general formula gives us exactly the same results which are the two answers separately. All at once, in order. Mathematicians did not deceive.)
Formulas for solving trigonometric equations with tangent and cotangent can also be checked. But let's not.) They are so unpretentious.
I painted all this substitution and verification on purpose. Here it is important to understand one simple thing: there are formulas for solving elementary trigonometric equations, just a summary of the answers. For this brevity, I had to insert plus/minus into the cosine solution and (-1) n into the sine solution.
These inserts do not interfere in any way in tasks where you just need to write down the answer to an elementary equation. But if you need to solve an inequality, or then you need to do something with the answer: select roots on an interval, check for ODZ, etc., these inserts can easily unsettle a person.
And what to do? Yes, either paint the answer in two series, or solve the equation / inequality in a trigonometric circle. Then these inserts disappear and life becomes easier.)
You can sum up.
To solve the simplest trigonometric equations, there are ready-made answer formulas. Four pieces. They are good for instantly writing the solution to an equation. For example, you need to solve the equations:
sinx = 0.3
Easily: x = (-1) n arcsin 0.3 + π n, n ∈ Z
cosx = 0.2
No problem: x = ± arccos 0.2 + 2π n, n ∈ Z
tgx = 1.2
Easily: x = arctg 1,2 + πn, n ∈ Z
ctgx = 3.7
One left: x= arcctg3,7 + πn, n ∈ Z
cos x = 1.8
If you, shining with knowledge, instantly write the answer:
x= ± arccos 1.8 + 2π n, n ∈ Z
then you already shine, this ... that ... from a puddle.) The correct answer is: there are no solutions. Don't understand why? Read what an arccosine is. In addition, if on the right side of the original equation there are tabular values \u200b\u200bof sine, cosine, tangent, cotangent, - 1; 0; √3; 1/2; √3/2 etc. - the answer through the arches will be unfinished. Arches must be converted to radians.
And if you already come across an inequality, like
then the answer is:
x πn, n ∈ Z
there is a rare nonsense, yes ...) Here it is necessary to trigonometric circle decide. What we will do in the corresponding topic.
For those who heroically read up to these lines. I just can't help but appreciate your titanic efforts. you a bonus.)
Bonus:
When writing formulas in an anxious combat situation, even hardened nerds often get confused where pn, And where 2πn. Here's a simple trick for you. In all formulas pn. Except for the only formula with arc cosine. It stands there 2πn. Two pien. Keyword - two. In the same single formula are two sign at the beginning. Plus and minus. Here and there - two.
So if you wrote two sign in front of the arc cosine, it is easier to remember what will happen at the end two pien. And vice versa happens. Skip the man sign ± , get to the end, write correctly two pien, yes, and catch it. Ahead of something two sign! The person will return to the beginning, but he will correct the mistake! Like this.)
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Trigonometric equations are not the easiest topic. Painfully they are diverse.) For example, these:
sin2x + cos3x = ctg5x
sin(5x+π /4) = ctg(2x-π /3)
sinx + cos2x + tg3x = ctg4x
Etc...
But these (and all other) trigonometric monsters have two common and obligatory features. First - you won't believe it - there are trigonometric functions in the equations.) Second: all expressions with x are within these same functions. And only there! If x appears somewhere outside, For example, sin2x + 3x = 3, this will be a mixed type equation. Such equations require an individual approach. Here we will not consider them.
We will not solve evil equations in this lesson either.) Here we will deal with the simplest trigonometric equations. Why? Yes, because the decision any trigonometric equations consists of two stages. At the first stage, the evil equation is reduced to a simple one by various transformations. On the second - this simplest equation is solved. No other way.
So, if you have problems in the second stage, the first stage does not make much sense.)
What do elementary trigonometric equations look like?
sinx = a
cosx = a
tgx = a
ctgx = a
Here a stands for any number. Any.
By the way, inside the function there may be not a pure x, but some kind of expression, such as:
cos(3x+π /3) = 1/2
etc. This complicates life, but does not affect the method of solving the trigonometric equation.
How to solve trigonometric equations?
Trigonometric equations can be solved in two ways. The first way: using logic and a trigonometric circle. We will explore this path here. The second way - using memory and formulas - will be considered in the next lesson.
The first way is clear, reliable, and hard to forget.) It is good for solving trigonometric equations, inequalities, and all sorts of tricky non-standard examples. Logic is stronger than memory!
We solve equations using a trigonometric circle.
We include elementary logic and the ability to use a trigonometric circle. Can't you!? However... It will be difficult for you in trigonometry...) But it doesn't matter. Take a look at the lessons "Trigonometric circle ...... What is it?" and "Counting angles on a trigonometric circle." Everything is simple there. Unlike textbooks...)
Ah, you know!? And even mastered "Practical work with a trigonometric circle"!? Accept congratulations. This topic will be close and understandable to you.) What is especially pleasing is that the trigonometric circle does not care which equation you solve. Sine, cosine, tangent, cotangent - everything is the same for him. The solution principle is the same.
So we take any elementary trigonometric equation. At least this:
cosx = 0.5
I need to find X. Speaking in human language, you need find the angle (x) whose cosine is 0.5.
How did we use the circle before? We drew a corner on it. In degrees or radians. And immediately seen trigonometric functions of this angle. Now let's do the opposite. Draw a cosine equal to 0.5 on the circle and immediately we'll see injection. It remains only to write down the answer.) Yes, yes!
We draw a circle and mark the cosine equal to 0.5. On the cosine axis, of course. Like this:
Now let's draw the angle that this cosine gives us. Hover your mouse over the picture (or touch the picture on a tablet), and see this same corner X.
Which angle has a cosine of 0.5?
x \u003d π / 3
cos 60°= cos( π /3) = 0,5
Some people will grunt skeptically, yes... They say, was it worth it to fence the circle, when everything is clear anyway... You can, of course, grunt...) But the fact is that this is an erroneous answer. Or rather, inadequate. Connoisseurs of the circle understand that there are still a whole bunch of angles that also give a cosine equal to 0.5.
If you turn the movable side OA for a full turn, point A will return to its original position. With the same cosine equal to 0.5. Those. the angle will change 360° or 2π radians, and cosine is not. The new angle 60° + 360° = 420° will also be a solution to our equation, because
Such full turns can be wound infinite set... And all these new angles will be solutions to our trigonometric equation. And they all need to be written down somehow. All. Otherwise, the decision is not considered, yes ...)
Mathematics can do this simply and elegantly. In one short answer, write down infinite set solutions. Here's what it looks like for our equation:
x = π /3 + 2π n, n ∈ Z
I will decipher. Still write meaningfully nicer than stupidly drawing some mysterious letters, right?)
π /3 is the same angle that we saw on the circle and identified according to the table of cosines.
2π is one full turn in radians.
n - this is the number of complete, i.e. whole revolutions. It is clear that n can be 0, ±1, ±2, ±3.... and so on. As indicated by the short entry:
n ∈ Z
n belongs ( ∈ ) to the set of integers ( Z ). By the way, instead of the letter n letters can be used k, m, t etc.
This notation means that you can take any integer n . At least -3, at least 0, at least +55. What do you want. If you plug that number into your answer entry, you get a specific angle, which is sure to be the solution to our harsh equation.)
Or, in other words, x \u003d π / 3 is the only root of an infinite set. To get all the other roots, it is enough to add any number of full turns to π / 3 ( n ) in radians. Those. 2πn radian.
Everything? No. I specifically stretch the pleasure. To remember better.) We received only a part of the answers to our equation. I will write this first part of the solution as follows:
x 1 = π /3 + 2π n, n ∈ Z
x 1 - not one root, it is a whole series of roots, written in short form.
But there are other angles that also give a cosine equal to 0.5!
Let's return to our picture, according to which we wrote down the answer. Here she is:
Move the mouse over the image and see another corner that also gives a cosine of 0.5. What do you think it equals? The triangles are the same... Yes! It is equal to the angle X , only plotted in the negative direction. This is the corner -X. But we have already calculated x. π /3 or 60°. Therefore, we can safely write:
x 2 \u003d - π / 3
And, of course, we add all the angles that are obtained through full turns:
x 2 = - π /3 + 2π n, n ∈ Z
That's all now.) In a trigonometric circle, we saw(who understands, of course)) all angles that give a cosine equal to 0.5. And wrote down these angles in brief mathematical form. The answer is two infinite series of roots:
x 1 = π /3 + 2π n, n ∈ Z
x 2 = - π /3 + 2π n, n ∈ Z
This is the correct answer.
Hope, general principle for solving trigonometric equations with the help of a circle is understandable. We mark on the circle the cosine (sine, tangent, cotangent) from given equation, draw the corners corresponding to it and write down the answer. Of course, you need to figure out what kind of corners we are saw on the circle. Sometimes it's not so obvious. Well, as I said, logic is required here.)
For example, let's analyze another trigonometric equation:
Please note that the number 0.5 is not the only possible number in the equations!) It's just more convenient for me to write it than roots and fractions.
We work according to the general principle. We draw a circle, mark (on the sine axis, of course!) 0.5. We draw at once all the angles corresponding to this sine. We get this picture:
Let's deal with the angle first. X in the first quarter. We recall the table of sines and determine the value of this angle. The matter is simple:
x \u003d π / 6
We recall full turns and, with a clear conscience, write down the first series of answers:
x 1 = π /6 + 2π n, n ∈ Z
Half the job is done. Now we need to define second corner... This is trickier than in cosines, yes ... But logic will save us! How to determine the second angle through x? Yes Easy! The triangles in the picture are the same, and the red corner X equal to the angle X . Only it is counted from the angle π in the negative direction. That's why it's red.) And for the answer, we need an angle measured correctly from the positive semiaxis OX, i.e. from an angle of 0 degrees.
Hover the cursor over the picture and see everything. I removed the first corner so as not to complicate the picture. The angle of interest to us (drawn in green) will be equal to:
π - x
x we know it π /6 . So the second angle will be:
π - π /6 = 5π /6
Again, we recall the addition of full revolutions and write down the second series of answers:
x 2 = 5π /6 + 2π n, n ∈ Z
That's all. A complete answer consists of two series of roots:
x 1 = π /6 + 2π n, n ∈ Z
x 2 = 5π /6 + 2π n, n ∈ Z
Equations with tangent and cotangent can be easily solved using the same general principle for solving trigonometric equations. Unless, of course, you know how to draw the tangent and cotangent on a trigonometric circle.
In the examples above, I used the tabular value of sine and cosine: 0.5. Those. one of those meanings that the student knows must. Now let's expand our capabilities to all other values. Decide, so decide!)
So, let's say we need to solve the following trigonometric equation:
There is no such value of the cosine in the short tables. Coldly ignore this creepy fact. We draw a circle, mark 2/3 on the cosine axis and draw the corresponding angles. We get this picture.
We understand, for starters, with an angle in the first quarter. To know what x is equal to, they would immediately write down the answer! We don't know... Failure!? Calm! Mathematics does not leave its own in trouble! She invented arc cosines for this case. Do not know? In vain. Find out. It's a lot easier than you think. According to this link, there is not a single tricky spell about "inverse trigonometric functions" ... It's superfluous in this topic.
If you're in the know, just say to yourself, "X is an angle whose cosine is 2/3." And immediately, purely by definition of the arccosine, we can write:
We remember about additional turns and calmly write down the first series of roots of our trigonometric equation:
x 1 = arccos 2/3 + 2π n, n ∈ Z
The second series of roots is also written almost automatically, for the second angle. Everything is the same, only x (arccos 2/3) will be with a minus:
x 2 = - arccos 2/3 + 2π n, n ∈ Z
And all things! This is the correct answer. Even easier than with tabular values. You don’t need to remember anything.) By the way, the most attentive will notice that this picture with the solution through the arc cosine essentially no different from the picture for cos equations x = 0.5.
Exactly! General principle that's why it's common! I specifically drew two almost identical pictures. The circle shows us the angle X by its cosine. It is a tabular cosine, or not - the circle does not know. What kind of angle is this, π / 3, or what kind of arc cosine is up to us to decide.
With a sine the same song. For example:
Again we draw a circle, mark the sine equal to 1/3, draw the corners. It turns out this picture:
And again the picture is almost the same as for the equation sinx = 0.5. Again we start from the corner in the first quarter. What is x equal to if its sine is 1/3? No problem!
So the first pack of roots is ready:
x 1 = arcsin 1/3 + 2π n, n ∈ Z
Let's take a look at the second angle. In the example with a table value of 0.5, it was equal to:
π - x
So here it will be exactly the same! Only x is different, arcsin 1/3. So what!? You can safely write the second pack of roots:
x 2 = π - arcsin 1/3 + 2π n, n ∈ Z
This is a completely correct answer. Although it does not look very familiar. But it's understandable, I hope.)
This is how trigonometric equations are solved using a circle. This path is clear and understandable. It is he who saves in trigonometric equations with the selection of roots on a given interval, in trigonometric inequalities- they are generally solved almost always in a circle. In short, in any tasks that are a little more complicated than standard ones.
Putting knowledge into practice?
Solve trigonometric equations:
At first it is simpler, directly on this lesson.
Now it's more difficult.
Hint: here you have to think about the circle. Personally.)
And now outwardly unpretentious ... They are also called special cases.
sinx = 0
sinx = 1
cosx = 0
cosx = -1
Hint: here you need to figure out in a circle where there are two series of answers, and where there is one ... And how to write down one instead of two series of answers. Yes, so that not a single root from an infinite number is lost!)
Well, quite simple):
sinx = 0,3
cosx = π
tgx = 1,2
ctgx = 3,7
Hint: here you need to know what is the arcsine, arccosine? What is arc tangent, arc tangent? Most simple definitions. But you don’t need to remember any tabular values!)
The answers are, of course, in disarray):
x 1= arcsin0,3 + 2πn, n ∈ Z
x 2= π - arcsin0.3 + 2
Not everything works out? It happens. Read the lesson again. Only thoughtfully(there is such an obsolete word...) And follow the links. The main links are about the circle. Without it in trigonometry - how to cross the road blindfolded. Sometimes it works.)
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You can order detailed solution your task!!!
Equality containing the unknown under the sign trigonometric function(`sin x, cos x, tg 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's write the root formulas for each of them.
1. Equation `sin x=a`.
For `|a|>1` it has no solutions.
With `|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 the sine, there are no solutions among real numbers.
With `|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`
It 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 sinus: 
For cosine: 
For tangent and cotangent: 
Formulas for solving equations containing inverse trigonometric functions:
Methods for solving trigonometric equations
The solution of any trigonometric equation consists of two stages:
- using to convert it to the simplest;
- solve the resulting simple equation using the above formulas for the roots and tables.
Let's consider the main methods of solution using examples.
algebraic method.
In this method, the replacement of a variable and its substitution into equality is done.
Example. Solve the equation: `2cos^2(x+\frac \pi 6)-3sin(\frac \pi 3 - x)+1=0`
`2cos^2(x+\frac \pi 6)-3cos(x+\frac \pi 6)+1=0`,
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`.
Decision. Move to the left all terms of equality: `sin x+cos x-1=0`. Using , we transform and factorize the left side:
`sin x - 2sin^2 x/2=0`,
`2sin x/2 cos x/2-2sin^2 x/2=0`,
`2sin x/2 (cos x/2-sin x/2)=0`,
- `sin x/2 =0`, `x/2 =\pi n`, `x_1=2\pi n`.
- `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 bring 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 split both parts by `cos x \ne 0` for the first case, and by `cos^2 x \ne 0` for the second. We get equations for `tg x`: `a tg x+b=0` and `a tg^2 x + b tg x +c =0`, which must be solved using known methods.
Example. Solve the equation: `2 sin^2 x+sin x cos x - cos^2 x=1`.
Decision. 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, dividing its left and right parts 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`, as a result `t^2 + t - 2=0`. The roots of this equation are `t_1=-2` and `t_2=1`. Then:
- `tg x=-2`, `x_1=arctg (-2)+\pi n`, `n \in Z`
- `tg x=1`, `x=arctg 1+\pi n`, `x_2=\pi/4+\pi n`, ` n \in Z`.
Answer. `x_1=arctg (-2)+\pi n`, `n \in Z`, `x_2=\pi/4+\pi n`, `n \in Z`.
Go to Half Corner
Example. Solve the equation: `11 sin x - 2 cos x = 10`.
Decision. Let's apply the formulas double angle, resulting in: `22 sin (x/2) cos (x/2) -` `2 cos^2 x/2 + 2 sin^2 x/2=` `10 sin^2 x/2+10 cos^ 2x/2`
`4 tg^2 x/2 - 11 tg x/2 +6=0`
Applying the above algebraic method, we get:
- `tg x/2=2`, `x_1=2 arctg 2+2\pi n`, `n \in Z`,
- `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 an auxiliary angle
In the trigonometric equation `a sin x + b cos x =c`, where a,b,c are coefficients and x is a variable, we divide both parts 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 1 and their modulus is at most 1. Let's 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`.
Decision. Dividing both sides of the equation 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`.
Denote `3/5 = cos \varphi` , `4/5=sin \varphi`. Since `sin \varphi>0`, `cos \varphi>0`, 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, in the numerators and denominators of which there are trigonometric functions.
Example. Solve the equation. `\frac (sin x)(1+cos x)=1-cos x`.
Decision. Multiply and divide the right side of the equation 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`
Given that the denominator cannot be zero, we get `1+cos x \ne 0`, `cos x \ne -1`, ` x \ne \pi+2\pi n, n \in Z`.
Equate the numerator of the fraction to zero: `sin x-sin^2 x=0`, `sin x(1-sin x)=0`. Then `sin x=0` or `1-sin x=0`.
- `sin x=0`, `x=\pi n`, `n \in Z`
- `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. The study begins in the 10th grade, there are always tasks for the exam, so try to remember all the formulas of trigonometric equations - they will definitely come in handy for you!
However, you don’t even need to memorize them, the main thing is to understand the essence, and be able to deduce. It's not as difficult as it seems. See for yourself by watching the video.
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The simplest trigonometric equations are the equations
Cos(x)=a, sin(x)=a, tg(x)=a, ctg(x)=a
Equation cos(x) = a
Explanation and rationale
- Roots cox equations= a. When | a | > 1 the equation has no roots because | cosx |< 1 для любого x (прямая y = а при а >1 or at a< -1 не пересекает график функцииy = cosx).
Let | a |< 1. Тогда прямая у = а пересекает график функции
y = cos x. On the interval, the function y = cos x decreases from 1 to -1. But a decreasing function takes each of its values only at one point of its domain of definition, therefore the equation cos x \u003d a has only one root on this interval, which, by definition of the arc cosine, is: x 1 \u003d arccos a (and for this root cos x \u003d a).
Cosine - even function, so on the interval [-n; 0] the equation cos x = and also has only one root - the number opposite to x 1, that is
x 2 = -arccos a.
Thus, on the interval [-n; n] (length 2n) the equation cos x = a for | a |< 1 имеет только корни x = ±arccos а.
The function y = cos x is periodic with a period of 2n, so all other roots differ from those found by 2np (n € Z). We get the following formula for the roots of the equation cos x = a when
x = ± arccos a + 2n, n £ Z.
- Particular cases of solving the equation cosx = a.
It is useful to remember the special notation for the roots of the equation cos x = a when
a \u003d 0, a \u003d -1, a \u003d 1, which can be easily obtained using the unit circle as a guide.
Since the cosine is equal to the abscissa of the corresponding point unit circle, we obtain that cos x = 0 if and only if the corresponding point of the unit circle is point A or point B.
Similarly, cos x = 1 if and only if the corresponding point of the unit circle is the point C, therefore,
x = 2πp, k € Z.
Also cos x \u003d -1 if and only if the corresponding point of the unit circle is the point D, thus x \u003d n + 2n,
Equation sin(x) = a
Explanation and rationale
- Roots sinx equations= a. When | a | > 1 the equation has no roots because | sinx |< 1 для любого x (прямая y = а на рисунке при а >1 or at a< -1 не пересекает график функции y = sinx).
