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Let real $b$, real $1 \geq a > 0$ and $0 \leq x,y,z \leq 2\pi$. Consider the three cyclic trig. equations:

$$ \sin(x) = a + b \sin(x-y) \\ \sin(y) = a + b \sin(y-z) \\ \sin(z) = a + b \sin(z-x) $$

Prove or disprove (if appropriate, under conditions for $a$ and $b$) that there is no other solution than $x=y=z=\arcsin(a)$.

I tried using the addition theorems for $\sin(x-y)$ etc. but still I cannot get conditions which support the claim.

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    As a simplest test whether this statement is true or not I think it would be reasonable to show whether non-diagonal solutions can branch from diagonal ones, excluding situation when $a = 1$ (two distinct solutions, $\pi - \arcsin a$ and $a$, merge into one). Such situations can be detected when Jacobi matrix of this system of equations is degenerate at solution for some parameter values. P.S. Can't get rid of an impression that this system describes equilibria of unidirectionally coupled ring of phase oscillators :) – 2017-01-17

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