What is the correct way to solve $\sin(2x)=\sin(x)$ 
I've found two different ways to solve this trigonometric equation

$\begin{align*}
\sin(2x)=\sin(x) \Leftrightarrow \\\\ 2\sin(x)\cos(x)=\sin(x)\Leftrightarrow \\\\ 2\sin(x)\cos(x)-\sin(x)=0 \Leftrightarrow\\\\ \sin(x) \left[2\cos(x)-1 \right]=0 \Leftrightarrow \\\\ \sin(x)=0 \vee \cos(x)=\frac{1}{2} \Leftrightarrow\\\\ x=k\pi \vee x=\frac{\pi}{3}+2k\pi \vee x=\frac{5\pi}{3}+2k\pi \space, \space k \in \mathbb{Z} 
\end{align*}$
The second way was:
$\begin{align*}
\sin(2x)=\sin(x)\Leftrightarrow \\\\ 2x=x+2k\pi \vee 2x=\pi-x+2k\pi\Leftrightarrow \\\\ x=2k\pi \vee3x=\pi +2k\pi\Leftrightarrow \\\\x=2k\pi \vee x=\frac{\pi}{3}+\frac{2k\pi}{3} \space ,\space k\in \mathbb{Z} 
\end{align*}$
What is the correct one?
Thanks

For what it's worth, the second one is "better" because it generalizes nicer. Imagine solving $\sin(3 x) = \sin(x)$ using the first method ($\sin(3 x) = 3\cos^2(x)\sin(x) - \sin^3(x)$). On the other hand, it's easy to see that $\sin(a x)= \sin(b x)$ will have an infinite number of solutions for any real $a$ and $b$ from the second method. 
