Is something wrong with this solution for $\sin 2x = \sin x$? I have this question. What are the solutions for $$ 
\sin 2x = \sin x;  \\ 0 \le x < 2 \pi $$
My method: 
$$ \sin 2x - \sin x = 0 $$
I apply the formula $$  \sin a - \sin b = 2\sin \left(\frac{a-b}{2} \right) \cos\left(\frac{a+b}{2} \right)$$
So:
$$ 2\sin\left(\frac{x}{2}\right) \cos\left(\frac{3x}{2}\right) = 0 $$
$$ \sin\left(\frac{x}{2}\right)\cos\left(\frac{3x}{2}\right) = 0 $$
Here one of the factors has to be $0$,
$$ \sin x = 0 \ \Rightarrow \  x = 0  \ or \ x = \pi $$
$$ \sin\left(\frac{x}{2}'\right) = 0 \ \Rightarrow \  x = 0  ;\  x \text{ can't be } \pi \text{ because of its range} $$
$$ \cos x = 0 \ \Rightarrow \  x = \frac{\pi}{2} \text{ or } \ x = \frac{3\pi}{2} $$
$$ \cos\left(\frac{3x}{2}\right) = 0 => x =  \frac{\pi}{3} \text{ or } x = \pi $$
So the solutions are : $$ 0, \pi,  \frac{\pi}{3} $$ 
I have seen other methods to solve this, so please don't post them. I'm really interested what's wrong with this one.

$\sin A=0\implies A=n\pi$
$\cos B=0\implies B=(2m+1)\pi/2$
$m,n$ are arbitrary integers
