Pre-Calculus trig

Question

Find the solutions to the following equations on the interval [0,2pi)

equations:

2 sin x + 2 = 0

2/3 sin^2 x - cos x = 1

Answer 1

2 sin x + 2 = 0

2 sin x = -2

six x = -1

x = -π/4, 3π/4

2/3 sin^2 x - cos x = 1

2sin²x - 3cosx = 3

2(1 - cos²x) -3cox -3 = 0

2 - 2cos²x - 3cosx -3 = 0

2cos²x+ 3cox + 1 = 0

(2cosx + 1)(cosx + 1) = 0

cosx = -1/2 cosx = -1

x = 5π/6 x = π