$\sin(x) = \sin(x − \frac\pi3)$, 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!
 A: 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 $x=-\frac{4\pi}{3}, -\frac{\pi}{3}, \frac{2\pi}{3}, \frac{5\pi}{3}$.
A: 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 $\frac53\pi$.
A: HINT: write $$\sin(x)-\sin(x-\pi/3)=0$$ this is equivalent to
$$\frac{1}{2}\left(\sin(x)+\sqrt{3}\cos(x)\right)=0$$
converting this into $\tan$ we get
$${\frac {\tan \left( x/2 \right) }{1+ \left( \tan \left( x/2 \right) 
 \right) ^{2}}}+1/2\,{\frac { \left( 1- \left( \tan \left( x/2
 \right)  \right) ^{2} \right) \sqrt {3}}{1+ \left( \tan \left( x/2
 \right)  \right) ^{2}}}
=0$$
set $$t=\tan(x/2)$$ and you can solve this equation.
simplifying your equation in $t$ you will get
$$1/6\,{\frac {\sqrt {3} \left( 3\,t+\sqrt {3} \right)  \left( -t+\sqrt 
{3} \right) }{{t}^{2}+1}}
=0$$
A: \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.
A: $$\sin x=\sin\left(x-\dfrac\pi3\right)$$
$$\implies x=m\pi+(-1)^m\left(x-\dfrac\pi3\right)$$ where $m$ is any integer
If $m$ is even $=2n$(say),
$$x=2n\pi+x-\dfrac\pi3\iff(6n-1)\dfrac\pi3=0\iff 6n=1$$ which is untenable as  $n$ is an integer
If $m$ is odd $=2n+1$(say)
$$x=(2n+1)\pi-\left(x-\dfrac\pi3\right)\iff x=\dfrac{\pi}3(2+3n)$$
We need $$-2\pi\le\dfrac{\pi}3(2+3n)\le2\pi\iff-6\le2+3n\le6$$
$2+3n\le6\iff n\le\dfrac43\implies n\le1$ as $n$ is an integer
$2+3n\ge-6\iff n\ge-\dfrac83\implies n\ge-2$ as $n$ is an integer
$\implies-2\le n\le1$ so allowing four possible values
