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.

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.

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.