(Should be) simple trigonometric equation: $\displaystyle \sin3x=\sin6x$.

Given solutions[1]$\displaystyle x=\frac{k\pi}{3}$,[2a]$\displaystyle x=\frac{\pi}{9}+\frac{2k\pi}{3}$,[2b]$\displaystyle x=-\frac{\pi}{9}+\frac{2k\pi}{3}$

Why is it, that with 'sum (difference) theorem' $\displaystyle \sin6x-\sin3x=0 \rightarrow 2\cos\frac{9x}{2}\sin\frac{3x}{2}=0$, I get different solutions than with $\displaystyle sin(3x) - sin(6x) = sin(3x) - 2sin(3x)cos(3x) = sin(3x)[1-2cos(3x)] = 0$ (at least it seems so)