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.

Other people have already pointed out we can rewrite like this:
$$2\sin(t)\cos(t) = \sin(t)$$
Now what most other people have not done, is this to substitute $\sin(t)$ and $\cos(t)$ for $y$ and $x$ and rewrite the problem to an algebraic variety:
$$\cases{2yx - y = 0\\
x^2+y^2=0}$$
This particular one is really silly easy to factor ( compared to systems of polynomial equations in general ):
$$\cases{y(2x-1) = 0\\x^2+y^2=0}$$
So either $2x-1 = 0$ or $y=0$
We recognize these equations as lines in the plane. The solutions are where they intercept the unit circle. Now it will be an easy exercise to substitute back and find all solutions.
