From Förberedande kurs i matematik 1

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

\displaystyle \sin 2x = \sin (-x)\,\textrm{.}

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

\displaystyle \sin u = \sin v

is satisfied if

\displaystyle u = v+2n\pi\qquad\text{or}\qquad u = \pi-v+2n\pi\,,

where n is an arbitrary integer. The consequence of this is that the solutions to the equation satisfy

\displaystyle 2x = -x+2n\pi\qquad\text{or}\qquad 2x = \pi-(-x)+2n\pi\,,

i.e.

\displaystyle 3x = 2n\pi\qquad\text{or}\qquad x = \pi +2n\pi\,\textrm{.}

The solutions to the equation are thus

\displaystyle \left\{\begin{align} x &= \frac{2n\pi}{3}\,,\\[5pt] x &= \pi + 2n\pi\,, \end{align}\right.

where n is an arbitrary integer.