$\sin(x) = \sin(x − \frac\pi3)$, solve for $x$ on interval $[-2\pi, 2\pi]$ According to the answer sheet:
$\sin(x) = \sin(x -\frac\pi3)$ gives:
$x = x-\frac\pi3 + k \cdot 2\pi$  or  $x = \pi-(x-\frac\pi3) + k \cdot 2\pi$
^ How did they go from $\sin(x) = \sin(x-\frac\pi3)$ to the equations above?
Thanks in advance!

There also exists a (seemingly) different approach. Recall that $$\sin x-\sin y =2\sin\frac{x-y}{2}\cos\frac{x+y}{2},$$
hence reforming your equation we obtain \begin{align}
0&=\sin x-\sin\left(x-\frac{\pi}{3}\right)\\&=2\sin\frac{x-\left(x-\frac{\pi}{3}\right)}{2}\cos\frac{x+\left(x-\frac{\pi}{3}\right)}{2}\\
&=2\sin\frac{\pi}{6}\cos\left(x-\frac{\pi}{6}\right)
\end{align} That is we have to find the roots of $\cos\left(x-\frac{\pi}{6}\right)$, which are where $$x-\frac{\pi}{6}=k\pi+\frac{\pi}{2}$$ or $$x=k\pi+\frac{2\pi}{3}$$
obtaining $x=-\frac43\pi,-\frac13\pi,\frac23\pi$ and $\frac53\pi$.
