Question

Find all solutions in the interval [0, 2π).

2 sin^2 x = sin x

Answer 1

0, π, 2π, π/6 and 5+(π/6)

Explanation:

Rearranging, sinx (2 sinx - 1) = 0.

In [0, 2π],sin x = 0 for x = 0, π and 2π.

The other factor = 0 gives the other two solutions.

sin 5+(π/6) = sin (π−(π-6)) = sin π/6 = 1/2.