How do you solve sin⁡2x=sin⁡x?

${\displaystyle \sin 2x=\sin x}$
${\displaystyle 2\sin x\cos x=\sin x}$
${\displaystyle 2\sin x\cos x-\sin x=0}$
${\displaystyle \sin x(2\cos x-1)=0}$
Solution A: ${\displaystyle \sin x=0\Rightarrow x=k\pi ,k\in \mathbb{Z}}$
Solution B: ${\displaystyle 2\cos x=1\Rightarrow \cos x=\frac{12}{,}x=\pm \frac{\pi }{3}+2k\pi =\frac{\pi }{3}+2k\pi =\frac{\pi }{3}(6k\pm 1),k\in \mathbb{Z}}$
${\displaystyle \therefore x=k\pi }$ or ${\displaystyle x=\frac{\pi }{3}(6k\pm 1),k\in \mathbb{Z}}$

Factor ${\displaystyle \sin \left(x\right)}$ out of ${\displaystyle 2\sin \left(x\right)\cos \left(x\right)-\sin \left(x\right)}$
${\displaystyle \sin \left(x\right)(2\cos \left(x\right)-1)=0}$
If any individual factor on the left side of the equation is equal to 0, the entire expression will be equal to 0.
${\displaystyle \sin \left(x\right)=0}$
${\displaystyle 2\cos \left(x\right)-1=0}$
Set the first factor equal to 0 and solve.
${\displaystyle x=2\pi n,\pi +2\pi n}$ for any integer n
Set the next factor equal to 0 and solve.
${\displaystyle x=\frac{\pi }{3}+2\pi n}$, for any integer n
The final solution is all the values that make ${\displaystyle \sin \left(x\right)(2\cos \left(x\right)-1)=0}$ true.
${\displaystyle x=2\pi n,\pi +2\pi n,\frac{\pi }{3}+2\pi n,\frac{5\pi }{3}+2\pi n}$, for any integer n
The finall solution is all the values that make ${\displaystyle \sin \left(x\right)(2\cos \left(x\right)-1)=0}$ true.
${\displaystyle x=2\pi n,\pi +2\pi n,\frac{\pi }{3}+2\pi n,\frac{5\pi }{3}+2\pi n}$, for any integer n
Consolidate ${\displaystyle 2\pi n}$ and ${\displaystyle \pi +2\pi n}$ to ${\displaystyle \pi n}$
${\displaystyle x=\pi n,\frac{\pi }{3}+2\pi n,\frac{5\pi }{3}+2\pi n}$ for any integer n

Very good answer