# 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?

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Even Wolfram is getting nowhere!! [1]:wolframalpha.com/input/… – Eisen Nov 8 '11 at 17:51
Might be a silly suggestion, but if you're just looking for a solution, $x = 0$ fits the bill. – Joel Cohen Nov 8 '11 at 18:26

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.

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This might be more illuminating: goo.gl/tVf4R – leonbloy Nov 8 '11 at 21:26
@leonbloy - you mean this: goo.gl/dxcxB The equation you have is not of the same form as the one from the original posting. Check the sin(x*y) vs. x*sin(y) – ja72 Nov 9 '11 at 16:06

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.

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