From Förberedande kurs i matematik 1

If we use the trigonometric relation \displaystyle \text{sin }\left( -x \right)=-\text{sin }x, the equation can be rewritten as

\displaystyle \sin 2x=\sin \left( -x \right)

In exercise 4.4:5a, we saw that an equality of the type

\displaystyle \sin u=\sin v\quad

is satisfied if

\displaystyle u=v+2n\pi or \displaystyle u=\pi -v+2n\pi

where \displaystyle n\text{ } is an arbitrary integer. The consequence of this is that the solutions to the equation satisfy

\displaystyle 2x=-x+2n\pi or \displaystyle 2x=\pi -\left( -x \right)+2n\pi

i.e.

\displaystyle 3x=2n\pi or \displaystyle x=\pi +2n\pi

The solutions to the equation are thus

\displaystyle \left\{ \begin{array}{*{35}l} x=\frac{2n\pi }{3} \\ x=\pi +2n\pi \\ \end{array} \right. ( \displaystyle n\text{ } an arbitrary integer)