Solve for x sin2x equals sin x, x is greater than 0 but less than 2 pi?

Question

Solve for x sin2x equals sin x, x is greater than 0 but less than 2 pi?

Answer 1

Sin 2x = 2 sin x cos x.

So 2 sin x cos x = sin x.

Thus

sin x = 0, so x = π, the only solution in (0, 2π),

or

2 cos x = 1

cos x = 1/2

x = π/3 or 5π/3.