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.

In restricting $0 \le x < 2\pi$ you are were inadvertently also restricting $0 \le 2x < 2\pi$ which is not a stated restriction.  
So $x = \pi$ is a solution as $\sin 2\pi$ does = $\sin \pi$ after all.
(Likewise $\sin 5\pi/3 = \sin 10\pi/3 = \sin 4\pi/3$).
So your restrictions are actually $0 \le 2x < 4\pi$ and for that matter $0 \le 3x/2 < 3\pi$.
So why did your answer not pick up $x = 5\pi/3$ as a potential answer.
As 
Then as $\cos(3x/2) = 0$ is potential answer, $0 \le 3x/2 < 3\pi$, so  $3x/2 = \pi/2, 3\pi/2, 5\pi/2$ so $x = \pi/3, \pi, 5\pi/3$.  
