Question

4(1+cosx)=3 sin^2x/2cosx/2

Answer 1

4(1+cosx)=3sin²x/2сosx/2
cos²x/2=1+cosx/2 |*2
2cos²x/2=1+cosx
4(2cos²x/2)=3sin²x/2*cosx/2
8cos²x/2-3sin²x/2*cosx/2=0
cosx/2(8cosx/2-3sin²x/2)=0
cosx/2=0
x/2=п/2+пn,n€z
x=п+2пn,n€z

8cosx/2-3sin²x/2=0
8cosx/2-3(1-cos²x/2)=0
3cos²x/2+8cosx/2-3=0
cosx/2=t
3t²+8t-3=0
D=64-4*3*(-3)=100
t1=-8+10/6=1/3
t2=-8-10/6=-18/6=-3
cosx/2=1/3
x/2=+-arccos1/3+2пn,n€z
x=+-2arccos1/3+4пn,n€z
cosx/2=-3
нет решения .
Ответ: x=п+2пn,n€z; x=+-2arccos1/3+4пn,n€z

Answer 2

First, we can simplify the right-hand side of the equation using the trigonometric identity:

sin^2x = (1 - cos2x)/2

cosx/2 = sqrt((1+cosx)/2)

Substituting these identities into the right-hand side of the equation, we get:

3 sin^2x/2cosx/2 = 3(1 - cos2x)/(2sqrt(1+cosx))

Next, we can simplify the left-hand side of the equation:

4(1+cosx) = 4cos^2(x/2)

Substituting this identity into the left-hand side of the equation, we get:

4cos^2(x/2) = 3(1 - cos2x)/(2sqrt(1+cosx))

Multiplying both sides by 2sqrt(1+cosx), we get:

8cos^2(x/2)sqrt(1+cosx) = 3(1 - cos2x)

Using the trigonometric identity:

cos2x = 2cos^2x - 1

We can simplify the right-hand side of the equation:

3(1 - cos2x) = 6cos^2x - 3

Substituting this identity into the equation, we get:

8cos^2(x/2)sqrt(1+cosx) = 6cos^2x - 3

Dividing both sides by 2cos^2(x/2), we get:

4sqrt(1+cosx) = 3sec^2(x/2)

Squaring both sides of the equation, we get:

16(1+cosx) = 9sec^4(x/2)

Using the trigonometric identity:

sec^2(x/2) = (1+cosx)/2

We can simplify the right-hand side of the equation:

9sec^4(x/2) = 9(1+cosx)^2/4

Substituting this identity into the equation, we get:

16(1+cosx) = 9(1+cosx)^2/4

Multiplying both sides by 4, we get:

64(1+cosx) = 9(1+cosx)^2

Expanding the square on the right-hand side of the equation, we get:

64(1+cosx) = 9(1+2cosx+cos^2x)

Simplifying the right-hand side of the equation, we get:

64(1+cosx) = 9 + 18cosx + 9cos^2x

Subtracting 64(1+cosx) from both sides of the equation, we get:

0 = 9cos^2x + 18cosx - 55

Using the quadratic formula, we can solve for cosx:

cosx = (-18 ± sqrt(18^2 - 4(9)(-55)))/(2(9))

cosx = (-18 ± sqrt(1080))/18

cosx = (-18 ± 6sqrt(30))/18

cosx = (-3 ± sqrt(30))/3

Now we can substitute these values of cosx back into the original equation to check if they satisfy the equation.