Question

Solve on the interval [0,2pi)

Answer 1

sin 2x = 2 sin x cos x

solve
sin 2x = - sin x
sin 2 x + sin x = 0
2 sin x cos x + sin x = 0
sin x( 2 cos x + 1) = 0
sin x = 0
x = n pi : n = 0, 1, 2

2 cos x + 1 = 0
cos x = -1/2
x = 2 pi/3 and 4 pi/ 3

ans:
x = 0, 2 pi/3, pi, 4 pi/3, 2 pi