Solve the following on the interval [0,2pie)

Question

Sin²x - sin x =0

Answer 1

sin^2x - sin x = 0

sin x (sin x - 1) = 0

sin x = 0, 1

On [0, 2π), sin x = 0 at 0 and π, and sin x = 1 at π/2

(0, π/2, π)