Find x such that sin2x=sinx & 0<=x<=2pi.

sin 2x  =  sin x      (using an identity on the left and simplifying)

2sinxcosx  - sin x  =  0            factor out sin x

sin x ( 2cos x  - 1)  =   0

We have two equations

sin x = 0    and this happens  at 0 , pi  and 2pi

2cos x - 1 =  0

2cos x  =  1

cos x =1/2    and this happens at pi/3  and  5pi/ 3

So the solutions are   0 , pi/3 , pi, 5pi/3  and 2 pi