Looking to find the analytic roots of $$(1+c) \cos(ax)+(1-c) \cos(bx)=0$$ where $a$,$b$, and $c>0$.

Any ideas? For $c=0$ the answer is straight forward.

P.S. The constants $a=A-B$ and $b=A+B$ if that helps better.

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    $\begingroup$ I'm not sure about solving this exactly, but I've simplified the equation to $$\cot Ax \times \cot Bx = C$$ (where $a=A-B$ and $b=A+B$, as mentioned above) $\endgroup$ – ExtremeRaider May 19 at 4:51
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    $\begingroup$ Further elaborating on my above comment: The equation can be modified into: $$\cos Ax \cos Bx - c \times \sin Ax \sin Bx =0.$$ For $c=1$, we end up with $$x=\frac{\pi}{2b} + 2n\pi [\forall n \in \mathbb N]$$ This is the most I could progress. I'm waiting for a proper generalization for all values of $c$. $\endgroup$ – ExtremeRaider May 19 at 4:58

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