Solving $\sin x = \sin 2x$ The equation is: $\sin x = \sin 2x$
I recognize that $\sin 2x$ is a double angle and use the double angle identity, so it becomes:
$$\begin{array}{rrcl}
&\sin x &=& 2 \sin  x \cos x  \\
\implies& \sin x - 2 \sin x \cos x &=& 0
\end{array}$$
Then I am stuck... Not sure how to proceed. 

$$\sin x = \sin 2x \implies \sin x = 2\sin x \cos x \implies \sin x - 2\sin x \cos x = 0 \implies \sin x (1-2\cos x) = 0$$
That means $\sin x = 0$ or $\cos x = \frac{1}{2}$
$\sin x = 0$ is true when $x = k\pi$
$\cos x = \frac{1}{2}$ is true when $x = \frac{\pm (1+ 2k\pi)}{3}$ or $x = \frac{\pm (5+ 2k\pi)}{3}$
So this statement holds when $x = k\pi$ or $x = \frac{\pm (1+ 2k\pi)}{3}$ or $x = \frac{\pm (5+ 2k\pi)}{3}$.
