Solve: cos2x+3sinx-2=0

Cos (2x) = 1 - 2sin^2x

cos (2x) + 3sin (x) - 2 = 0

1 - 2sin^2 (x) + 3sin (x) - 2 = 0

-2sin^2 (x) + 3sin (x) - 1 = 0

2sin^2 (x) - 3sin (x) + 1 = 0

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

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

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

sin (x) = {1/2, 1}

In the range [0, 2π], x = {π/6, π/2, 5π/6}