Trig quadratics

Question

Find all solutions of 2sin^ x + cos x - 2 = 0, where x is greater than or equal to 0 and less than or equal to 2pi

Answer 1

2sin²x + cosx - 2 = 0

2(1 - cos²x) + cosx - 2 = 0

-2cos²x + cosx = 0

cosx(-2cosx + 1) = 0

cosx = 0 or cosx = 1/2

cosx = 0 when x = π/2, 3π/2

cosx = 1/2 when x = π/3, 5π/3