Solution of $\left|\frac{\sin Mx}{\sin x}\right| = \left|\frac{\sin NMx}{\sin Nx}\right|$ I am looking for the smallest solution of 
$$\left|\frac{\sin Mx}{\sin x}\right| = \left|\frac{\sin NMx}{\sin Nx}\right|$$
where $M < N - 1$ are integers and $x \in \mathbb{R}^+$.
By plotting both sides for different $N$ and $M$'s I noticed that the smallest solution is independent of $M$. In other words, for a given $N$ changing $M$ doesn't result in the change of the smallest solution. I can't figure out this independence.
For example, $\left|\frac{\sin Mx}{\sin x}\right| - \left|\frac{\sin NMx}{\sin Nx}\right|$ is plotted for $N = 7$ and different values of $M$ in the figure below.

 A: For integer $k$ we have $\sin(k\pi + x) = (-1)^{k}\sin(x) \implies |\sin(k\pi+x)| = |\sin(x)|$ so if $x = \frac{\pi}{N+1}$ it follows that $$|\sin(Nx)| = |\sin(x)|, ~~~|\sin(Mx)| = |\sin(MNx)|\implies \left|\frac{\sin(Mx)}{\sin(x)}\right| = \left|\frac{\sin(MNx)}{\sin(Nx)}\right|$$
Thus $x = \frac{\pi}{N+1}$ is a solution independent of $M$. I suspect this is the solution you have observed. Numerically one can check that this is the smallest positive solution for the first few values of $N$ if $M<N-1$ (I checked it for $N<20$), but proving this seems messy.
By symmetry and periodicy we also have the ($M$-independent) solutions $x = -\frac{\pi}{N+1}$, $x=0$, $x = k\pi$ and $x = k\pi \pm \frac{\pi}{N+1}$ for any integer $k$.
A: Forget about the absolute value and transform the difference $$f(N,M)=\frac{\sin Mx}{\sin x}- \frac{\sin NMx}{\sin Nx}$$
For example if $M=2$ and $N=7$ then I claim that: $$f(7,2)=2(\cos x - \cos 7x)$$
Generalize and this will give you the idea when the root is independent of M.
