Question

Sin^2(7п/2+х)-sin2x=0​

Answer 1

Ответ:

Пошаговое объяснение:

2*sinx*cosx=cosx

s*sinx*cosx-cosx=0

cosx*(2*sinx-1)=0

cosx=0 или sinx=1/2

x=Pi/2+Pi*n;

x=((-1)^m)*(7*Pi/6)+Pi*m

-7*Pi/2;-17*Pi/6

Answer 2

To solve the equation sin^2(7π/2 + x) - sin(2x) = 0, you can follow these steps:

1. Use the trigonometric identity sin^2(θ) = (1 - cos(2θ))/2 to rewrite sin^2(7π/2 + x):

sin^2(7π/2 + x) = (1 - cos(2(7π/2 + x)))/2

2. Simplify the expression inside the cosine:

cos(2(7π/2 + x)) = cos(7π + 2x)

Since cos(7π) = -1 and cos(2x) = cos(-2x), the expression becomes:

-1 * cos(-2x) = cos(-2x)

3. Substitute this simplified expression back into the original equation:

(1 - cos(-2x))/2 - sin(2x) = 0

4. Multiply both sides of the equation by 2 to get rid of the fraction:

1 - cos(-2x) - 2sin(2x) = 0

5. Rearrange the terms:

1 - 2sin(2x) - cos(-2x) = 0

6. Use the identity cos(-θ) = cos(θ) to simplify the equation:

1 - 2sin(2x) - cos(2x) = 0

7. Now, you have a quadratic equation in terms of sin(2x):

2sin(2x) + cos(2x) - 1 = 0

8. To solve for sin(2x), you can use a substitution. Let y = sin(2x), so the equation becomes:

2y + cos(2x) - 1 = 0

9. Solve for y:

2y + cos(2x) = 1

2y = 1 - cos(2x)

y = (1 - cos(2x))/2

10. Now that you have found y, substitute sin(2x) back in:

sin(2x) = (1 - cos(2x))/2

11. To solve for x, you can use trigonometric identities. First, square both sides:

sin^2(2x) = (1 - cos(2x))^2/4

12. Now, use the identity sin^2(θ) = 1 - cos^2(θ) to rewrite sin^2(2x):

1 - cos^2(2x) = (1 - cos(2x))^2/4

13. Multiply both sides by 4 to eliminate the fraction:

4 - 4cos^2(2x) = (1 - cos(2x))^2

14. Expand the right side:

4 - 4cos^2(2x) = 1 - 2cos(2x) + cos^2(2x)

15. Rearrange the terms and simplify:

4cos^2(2x) - 2cos(2x) - 3 = 0

16. Now, you have a quadratic equation in terms of cos(2x). You can solve for cos(2x) using the quadratic formula:

cos(2x) = [2 ± √(2^2 - 4 * 4 * (-3))] / (2 * 4)

cos(2x) = [2 ± √(4 + 48)] / 8

cos(2x) = [2 ± √52] / 8

cos(2x) = (1 ± √13) / 4

17. Now, you can find the values of 2x by taking the inverse cosine of the solutions:

2x = cos^(-1)((1 + √13)/4) and 2x = cos^(-1)((1 - √13)/4)

18. Finally, solve for x by dividing both sides by 2:

x = (1/2)cos^(-1)((1 + √13)/4) and x = (1/2)cos^(-1)((1 - √13)/4)

These are the solutions for the equation sin^2(7π/2 + x) - sin(2x) = 0 in terms of x.