Since sin(2x) = 2 sinx cosx, we have
2 sinx cosx sinx - cosx = 0
cosx (2 sin²x - 1) = 0
Therefore
cosx = 0 or 2 sin²x - 1 = 0
The solutions of cosx = 0 are x = any whole number of rightangles.
For 2 sin²x - 1, we get
sinx = 1/√2 or -1/√2
so x = any odd multiple of a half-rightangle, i.e. 45°, 135°, 225° etc, or, in radians,
x = π/4, 3π/4, 5π/4 etc