What are the solutions to the trigonometric equation on the interval [0,2π) ?

I think this is supposed to be

2sin^2 x  = sin x

2*sin^2 x - sin x  =  0              factor out sin x

sin x ( 2sin x - 1)  =  0

sin x  = 0       and this occurs when x = 0   and x = pi

And

2sin x - 1  = 0

2sin x  = 1

sin x  =1/2    and this occurs when x = pi/6  and x  = 5pi/6