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

You have all of the right ingredients.
The equation $\sin(x) = 0$ yields solutions $x = k\pi$ for $k\in \mathbb{Z}$.
The equation $\cos(x) = 0$ yields solutions $x = \pi/3, 5\pi/3$ in $[0,2\pi]$.
Since the cosine function is $2\pi$-periodic, you get the solutions 
$$x = \pi(1/3 + 2k), \pi(5/3 + 2k), \qquad k\in\mathbb{Z}.$$
So your solution is the totality of all of these.
