what are all the exact solutions of 2 sin^2 x- sin x = 0

Factor:

2sin^2x-sinx = 0

sinx(2sinx - 1) = 0

Therefore the solutions are when:

sin x = 0
And
sinx = 1/2

So sinx = 0
is true when x = 0 and pi  and all the angles coterminal with these points.  Thus, the answer is x = pi*n, (where n is some integer)

sinx = 1/2
is true when x = pi/6 and 5pi/6 and the angles coterminal with these points.
Thus, the answer is x = pi/6 + 2pi*n (where n is some integer)
and x = 5pi/6 + 2pi*n (where n is some integer)