Finding all solutions of $2\sin(c_1\omega)+\sin(c_2\omega)=0$ for $\omega$, when $c_1$ and $c_2$ are given positive real numbers?

Question

I try to solve the following problem, which consists of 1 equation and 1 unknown. \begin{align} 2\sin(c_1\omega)+\sin(c_2\omega)=0 \end{align} Suppose $c_1$ and $c_2$ are known positive real numbers. $\omega$ is the variable which needs to be solved for. $\omega=0$ is one solution but is there a way to get a couple more solutions in the positive direction? I tried to bring it in a polynomial form. Let $\omega = \cos^{-1}(z)$, then: \begin{align} 2\sin(c_1\cos^{-1}(z))+\sin(c_2\cos^{-1}(z))=0 \end{align} If $c_1$ and $c_2$ would be just integers I could simplify it much easier but this is not the case. Is there anyway to rewrite this in a different way, and maybe extract $\omega$? Any help or hint is appreciated. Thank you.

Answer 1

COMMENT.- $(1)$ Plotting the function $f(\omega)=2\sin(c_1\omega)+\sin(c_2\omega)$ with $c_1,c_2$ positive real, one can see the graphic can be very complicated. However it seems $f(\omega)$ is always periodic. It follows that if you can find out the period in each case your are done because $f(0)=0$.

Unfortunately the determination of these periods are not at all easy. For example for $(c_1,c_2)=(1.3,3.7)$ Wolfram gives the period $20\pi$ and for $(c_1,c_2)=(5.2,2.3)$ him gives the period $62.8319$ which is almost $20\pi$. I am not able to calculate these two periods and less to calculate the periods corresponding to literal values of $c_1$ and $c_2$.

$(2)$ Another way to look at this problem could be the analisis of the rational function $F(X,Y)$ defined by $$F(X;Y)= \frac{2(2X)}{X^2+1}+\frac{2Y}{Y^2+1}$$ where $X=\tan\left(\dfrac{c_1\omega}{2}\right)$ and $Y=\tan\left(\dfrac{c_2\omega}{2}\right)$.

But this is a simple COMMENT.