TITLE: Finding roots of $\sin(x)=\sin(ax)$ without resorting to complex analysis
QUESTION [6 upvotes]: If you were given an equation $\sin(x)=\sin(ax)$ (say $a$ is a natural number), how would you go about finding all the roots on $[0,2\pi)$ without delving into complex numbers? 
From a simple geometric analysis it is obvious that solving for $\pi-x=ax$ would yield 4 solutions, and $x=0$ and $x=\pi$ are another 2 obvious solutions. From complex analysis though we know that there could be many roots in this interval depending on the $a$. Is there any way to find all these roots using only techniques from real analysis?

REPLY [8 votes]: $$\sin(x) - \sin(ax) = 2\sin\left(\frac{x-ax}{2}\right)\cos\left(\frac{x+ax}{2}\right) = 0 $$  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.