I have been attempting to solve the following equation for $\theta $ ($\gamma$ is a real number, $n$ is a natural number)
$sin((n+1)\theta) +\gamma sin(n\theta)=0$
without much success. It seems like the solution should not be too difficult, but the answer still eludes me. Am I missing something obvious here? Any help would be appreciated.
I don't think this is an easily solvable problem. For example, taking $n=3$ and $\gamma=2$, and using $\sin t=(e^{it}-e^{-it})/2i$, yields a degree $8$ polynomial in $e^{it}$ that needs to be solved: $(x-1)(x+1)(x^6+2 x^5+x^4+2 x^3+x^2+2 x+1)=0$. The degree $6$ factor is irreducible and has four roots on the unit circle, which cannot be expressed in radicals. In this particular case, the imaginary parts of the roots can be expressed in radicals, so you could write the answer as arcsin of something complicated. But in general, I don't see why it should be possible to find a closed form for the solution (unless $\gamma=\pm1$ or something).