Cos 1 are special cases. Trigonometric Equations - Formulas, Solutions, Examples
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The video course "Get an A" includes all the topics necessary for a successful passing the exam in mathematics for 60-65 points. Completely all tasks 1-13 profile exam mathematics. Also suitable for passing the Basic USE in mathematics. If you want to pass the exam with 90-100 points, you need to solve part 1 in 30 minutes and without mistakes!
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The main methods for solving trigonometric equations are: reducing equations to the simplest ones (using trigonometric formulas), introduction of new variables, factorization. Let's consider their application with examples. Pay attention to the registration of the solution of trigonometric equations.
A necessary condition for the successful solution of trigonometric equations is the knowledge of trigonometric formulas (topic 13 of work 6).
Examples.
1. Equations Reducing to the Simplest.
1) Solve the equation
Solution:
Answer:
2) Find the roots of the equation
(sinx + cosx) 2 = 1 – sinxcosx belonging to the segment .
Solution:
Answer:
2. Equations Reducing to Quadratic Equations.
1) Solve the equation 2 sin 2 x - cosx -1 = 0.
Solution: Using sin formula 2 x \u003d 1 - cos 2 x, we get
Answer:
2) Solve the equation cos 2x = 1 + 4 cosx.
Solution: Using the formula cos 2x = 2 cos 2 x - 1, we get
Answer:
3) Solve the equation tgx - 2ctgx + 1 = 0
Solution:
Answer:
3. Homogeneous equations
1) Solve the equation 2sinx - 3cosx = 0
Solution: Let cosx = 0, then 2sinx = 0 and sinx = 0 - a contradiction with the fact that sin 2 x + cos 2 x = 1. So cosx ≠ 0 and you can divide the equation by cosx. Get
Answer:
2) Solve the equation 1 + 7 cos 2 x = 3 sin 2x
Solution:
Using the formulas 1 = sin 2 x + cos 2 x and sin 2x = 2 sinxcosx, we get
sin2x + cos2x + 7cos2x = 6sinxcosx
sin2x - 6sinxcosx+ 8cos2x = 0
Let cosx = 0, then sin 2 x = 0 and sinx = 0 - a contradiction with the fact that sin 2 x + cos 2 x = 1.
So cosx ≠ 0 and we can divide the equation by cos 2 x .
Get
tg 2x – 6 tgx + 8 = 0
Denote tgx = y
y 2 – 6 y + 8 = 0
y 1 = 4; y2=2
a) tanx = 4, x= arctg4 + 2 k, k
b) tgx = 2, x= arctg2 + 2 k, k .
Answer: arctg4 + 2 k, arctan2 + 2 k, k
4. Equations of the form a sinx + b cosx = with, with≠ 0.
1) Solve the equation.
Solution:
Answer:
5. Equations Solved by Factorization.
1) Solve the equation sin2x - sinx = 0.
The root of the equation f (X) = φ ( X) can only serve as the number 0. Let's check this:
cos 0 = 0 + 1 - the equality is true.
The number 0 is the only root of this equation.
Answer: 0.
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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`.
Solution. 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 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`.
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, dividing 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`, 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`.
Solution. Applying the double angle formulas, the result is: `22 sin (x/2) cos (x/2) -` `2 cos^2 x/2 + 2 sin^2 x/2=` `10 sin^2 x/2 +10 cos^2 x/2`
`4 tg^2 x/2 - 11 tg x/2 +6=0`
Applying the above algebraic method, we get:
- `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 equal to 1 and their modulus is not greater than 1. 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. 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`.
Solution. 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.
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
- The roots of the equation cosx = 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 on the unit circle, we get that cos x = 0 if and only if the corresponding point on 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
- The roots of the equation sinx = 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).
