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.

There's nothing wrong up to the reduction to
$$
\sin\frac{x}{2}\cos\frac{3x}{2}=0
$$
Then you have either
$$
\sin\frac{x}{2}=0
$$
that is, $x/2=k\pi$ and $x=2k\pi$, or
$$
\cos\frac{3x}{2}=0
$$
so
$$
\frac{3x}{2}=\frac{\pi}{2}+k\pi
$$
and
$$
x=\frac{\pi}{3}+\frac{2k\pi}{3}
$$
Now let's examine the first set of solutions; you want to find the integers $k$ such that
$$
0\le 2k\pi<2\pi
$$
and this is only for $k=0$. For the second set of solutions,
$$
0\le\frac{\pi}{3}+\frac{2k\pi}{3}<2\pi
$$
becomes
$$
0\le 1+2k<6
$$
which gives $k\in\{0,1,2\}$.
Thus you find
$$
x\in\left\{0,\frac{\pi}{3},\pi,\frac{5\pi}{3}\right\}
$$
