math help needed

Question

how to solve 2cos²x - 3sinx = 3

restriction is π/2 ≤ x ≤ 3π/2

Answer 1

2 (1-sin^2) - 3sin = 3

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

2 - 2sin^2 - 3 sin - 3 = 0

-2sin^2 - 3 sin - 1 = 0

2 sin^2 + 3 sin + 1 = 0

2 M^2 + 3M + 1 = 0

(2M + 1 )(M + 1) = 0

M = -1/2 M = -1

sin = -1/2 sin = -1

sine is -1/2 at 210 and 330

sine is -1 at 270

but we're only good from 90 to 270

so 210 and 270 or 7 pi / 6 and 3 pi/2

you're welcome