<s>[INST] $\sin(x) = \sin(x − \frac\pi3)$, 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!
 [/INST] The first comes from the fact that $\sin(x) = \sin(x \pm 2\pi n) \,\, n\in \mathbb{Z}$, because $\sin$ has period $2\pi$ The second comes from $\sin(x) = \sin(\pi - x)$. 
To answer the question in your title, however, about the intersections on the interval $[-2\pi, 2\pi]$, we'll reduce this to a formula for $x$. 
\begin{align*}
\sin(x) &= -\cos(x + \pi/6) \\
1/2 (\sqrt{3} \cos x + \sin x) &= 0 \\
\tan x &= -\sqrt{3} \\
x &= \pi n - \frac{\pi}{3} \,\, n\in\mathbb{Z}
\end{align*}
which has values on $[-2\pi, 2\pi]$ of $x=-\frac{4\pi}{3}, -\frac{\pi}{3}, \frac{2\pi}{3}, \frac{5\pi}{3}$.
</s>