Finding roots of $\sin(x)=\sin(ax)$ without resorting to complex analysis

$$\sin(x) - \sin(ax) = 2\sin\left(\frac{x-ax}{2}\right)\cos\left(\frac{x+ax}{2}\right) = 0 $$ This identity is not difficult to prove.

Solving $\frac{x-ax}{2} = {\pi}n$ and $\frac{x+ax}{2} = \frac {\pi}{2} + \pi{n}$ so that

$$x=\frac{2\pi{n}}{1-a} , \frac{2\pi{n} + \pi}{1+a}$$ for $a \neq 1, -1$. The case $a=1$ is trivial, and $a=-1$ yields solutions whenever $\sin$ vanishes, namely $\pi{n}$ which is clear since $\sin$ is an odd function.