Solving $\frac{\sin(\alpha \cdot x)}{\alpha}=\frac{\sin(\beta \cdot x)}{\beta}$ I am looking for a solution for the equation
$$\frac{\sin(\alpha \cdot x)}{\alpha}=\frac{\sin(\beta \cdot x)}{\beta}$$
where $\alpha$ and $\beta$ are constants.
How do I approach this? 
 A: Flip it around and plot contour lines of $\alpha$ vs. $\beta$ for various values of $x$. Ignoring the obvious $\alpha=\beta$ solutions, the resulting plot is a real beauty.

A: This answer assumes you are interested in the solutions of this equaton in the real line. 
Using $u=\alpha x$ and $\gamma=\beta/\alpha$, the question is equivalent to finding the zeroes of the function $g$ defined by
$$
g(u)=\gamma\sin(u)-\sin(\gamma u).
$$
Assume without loss of generality that $\gamma>1$. Then, $\sin(u)=1$ implies $g(u)\geqslant\gamma-1>0$ and $\sin(u)=-1$ implies $g(u)\leqslant-\gamma+1<0$, hence $g$ has at least one zero in each interval $I_k$ of length $\pi$ around $k\pi$, for $k$ in $\mathbb Z$. A zero in $I_0$ is $u=0$. 
There might be exactly one zero in each interval $I_k$. This is so in the simulations I performed but these are based on a very limited number of parameters $\gamma$ and of intervals $I_k$.
If $\gamma$ is a rational number $\gamma=m/n$ with $m$ and $n$ integers, then $2n\pi$ is a period of $g$ hence the zeroes in $I_{k+2n}$ are exactly those in $I_k$ translated by $2n\pi$. If $\gamma$ is irrational, I see no reason to suspect a similar regularity.
