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

These answers are equivalent and both are correct. Placing angle $x$ on a unit circle, your first decomposition gives all angles at the far west and east sides, then all the angles $60$ degrees north of east, then all the angles $60$ degrees south of east.
Your second decomposition takes all angles at the far east side first. Then it takes all angles spaced  one-third around the circle starting at 60 degrees north of east. You have the same solution set either way.
