find all solutions between 0<x<2pi

Question

2cos^2(x) = 3sin(x)+3

Answer 1

2 cos² x = 3 sin x + 3

2 (1 - sin² x) = 3 sin x + 3

2 - 2 sin² x = 3 sin x + 3

0 = 2 sin² x + 3 sin x + 1)

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

Can you continue and answer?

Answer 2

You would like everything in terms of sin. Thus,

2cos^2(x) = 3sin(x)+3

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

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

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

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

x = 7π/6, 3π/2, 11π/6