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. 
 A: \begin{align}
\sin x & = \sin 2x \\
&= 2 \sin x \cos x
\end{align}
Then 
\begin{align}
\sin x - 2 \sin x \cos x = 0
\end{align}
So
$$\sin x (1 - 2\cos x ) = 0$$
This imples that $\sin x = 0$ or $(1-2\cos x) = 0$.
$\sin x = 0$ implies that $x = n \pi$. Also $(1 - 2\cos x ) = 0$ implies that $\cos x = \frac{1}{2}$ and so $x = 2n \pi + \frac{5\pi}{3}, 2n \pi + \frac{\pi}{3}$. So that the solution set is $$\left \{n \pi,2n \pi + \frac{5\pi}{3}, 2n \pi + \frac{\pi}{3}: n \in\mathbb{Z}\right\}$$
A: Hint:
In general if $\sin x=\sin y,$
$x=n\pi+(-1)^ny$ where $n$ is any integer.
Check for $n$ is even and $n$ is odd one by one
A: $$\sin x = \sin 2x \implies \sin x = 2\sin x \cos x \implies \sin x - 2\sin x \cos x = 0 \implies \sin x (1-2\cos x) = 0$$
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 $x = \frac{\pm (5+ 2k\pi)}{3}$.
