<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] \begin{eqnarray}
\sin x&=&\sin\left(x-\frac{\pi}{3}\right)\\
&=&\sin x\cos\frac{\pi}{3}-\cos x\sin\frac{\pi}{3}\\
&=&\frac{1}{2}\sin x-\frac{\sqrt{3}}{2}\cos x\\
\frac{1}{2}\sin x&=&-\frac{\sqrt{3}}{2}\cos x\\
\tan x&=&-\sqrt{3}\tag{1}\\
x&=&-\frac{\pi}{3},-\frac{4\pi}{3},\frac{2\pi}{3},\frac{5\pi}{3}
\end{eqnarray}
This answer the title of your question.
To answer the second question contained within the text of your question, once one has equation $(1)$ above, one knows that the general solution will be
$$ x=\tan^{-1}\left(-\sqrt{3}\right)+\pi k=-\frac{\pi}{3}+\pi k$$
which is equivalent to
$$ x=\frac{2\pi}{3}+\pi k$$
As for the results which you give, see the following:
Using the identities
$$ \sin x=\sin(\pi-x) $$
and
$$\sin x=\sin(x+2\pi k)$$
if $\sin(x)=\sin\left(x-\frac{\pi}{3}\right)$ then either
$$ x=x-\frac{\pi}{3}+2\pi k\quad\text{ for some integer }k\tag{2} $$
or
$$ \pi-x=x-\frac{\pi}{3}+2\pi k\quad\text{ for some integer }k\tag{3} $$
Equation $(2)$ gives the nonsensical result $k=\frac{1}{6}$ and equation $(3)$ gives
$$x=\frac{2\pi}{3}-\pi k$$
which is equivalent to
$$x=\frac{2\pi}{3}+\pi k$$
COMMENT: Your answer sheet did the same thing I did in equation $(2)$ and $(3)$ except they replaced the $x-\frac{\pi}{3}$ on the right side of the equation with $\pi-(x-\pi/3)$ whereas I replaced the $x$ on the left side with $\pi-x$ which I think is a little less confusing.
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