Is there any analytical solution of $\sin mx =k \sin x$ for $x$, where $m>1$ and $k \neq 0$?

Question

Is there any analytical solution of $\sin mx =k \sin x$ for $x$, where $m>1$ and $k\neq 0$? I also need solutions of $\cos mx =k \cos x$ and $\tan mx =k \tan x$ for $x$. When $k=1$, the solution is $x=\pi/(1+m)$. But cannot solve for $k \neq 1$.

Answer 1

Of course $x = 0$ is always a solution. Similarly, if $m$ is an integer, $x = $ any multiple of $\pi$ is a solution. But I suppose you don't want those.

If $m$ is a positive integer, $\sin(mx) = \sin(x) U_{m-1}(\cos(x))$, where $U_j$ are the Chebyshev polynomials of the second kind. So solutions are $x = 2 n \pi \pm \arccos(r)$ where $r$ is a root of $U_{m-1}(r) - k$ and $n$ is any integer.