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.

If we need to solve $$\sin \frac{x}{2} \cos \frac{3x}{2} = 0$$ for $0 \le x < 2\pi$, then the inequality condition is equivalent to $$0 \le x/2 < \pi,$$ and $$0 \le 3x/2 \le 3\pi.$$  Thus $\sin x/2 = 0$ admits only $x = 0$ in first interval; and in the second interval, we have solutions $$\frac{3x}{2} \in \left\{ \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2} \right\},$$ as all of these are between $0$ and $3\pi$.  Consequently, the second factor admits the solutions $$x \in \left\{\frac{\pi}{3}, \pi, \frac{5\pi}{3} \right\}.$$ The complete solution set is therefore $$x \in \left\{0, \frac{\pi}{3}, \pi, \frac{5\pi}{3} \right\}$$ in the desired interval.

A plot of the curves $$\color{blue}{y = \sin 2x}, \quad \color{orange}{y = \sin x}, \quad \color{green}{y = \sin 2x - \sin x}$$ is shown below:

As you can see, the green curve intersects the $x$-axis at the claimed values (as well as at $2\pi$ but this was excluded by the condition $0 \le x < 2\pi$).  You can also visually see that the $x$-coordinates at which the blue and orange curves coincide (corresponding to the equality $\sin 2x = \sin x$) are also the $x$-coordinates at which the green curve intersects the $x$-axis.
