Solve for all solutions on interval 0 2 pi sin 2x sin x 0S

Solve for all solutions on interval [0, 2 pi). sin (2x) + sin (x) = 0

Solution

sin(2x) + sinx = 0
2sin(x)cos(x) + sin(x) = 0
sin(x)[2cos(x) + 1] = 0
sin(x) = 0 or 2cos(x) + 1 = 0
sin(x) = 0 or cos(x) = –1/2
sin(x) = 0 x = k
cos(x) = –1/2 x = ±2/3 + 2k
(k is any integer)

If x belongs to [0,2), then the solution set will be {0, 2/3, , 4/3}