Solve the following equations for all solutions of x

2sin^2x+3cos-3=0

2 sin² x + 3 cos x - 3 = 0

2 (1 - cos² x) + 3 x - 3 = 0

2 - 2 cos² x + 3 cos x - 3 = 0

- 2 cos² x + 3 cos x - 1 = 0 / * ( - 1)

2 cos² x - 3 cos x + 1 = 0

Substitution: u = cos x

2 u² - 3 u + 1 = 0

2 u² - 2 u - u + 1 = 0

2 u (u - 1) - (u - 1) = 0

(u - 1) (2 u - 1) = 0, u 1 = 1, u 2 = 1/2

cos x = 1, cos x = 1/2

Answer:

x 1 = k π, x 2 = π / 3 + 2 k π, x 3 = 5 π / 3 + 2 k π, k ∈ Z

Remember that sin^2x + cos^2x = 1 so sin^2x = 1 - cos^2x

2sin^2 x + 3cosx - 3 = 0

2 (1 - cos^2x) + 3cosx - 3 = 0

2cos^2x - 3cosx + 1 = 0

This is just a quadratic equation in cosx

So let u = cosx

2u^2 - 3u + 1 = 0

This factors as

(2u-1) (u-1) = 0

So 2u = 1 and u = 1 or 2cosx = 1 and cosx = 1

cosx = 1 when x = 0cosx = 1/2 when x = pi/3 and 5pi/3

So x = {0, pi/3, 5pi/3} is the solution set for your equation.