# Finding roots for $(1+c) \cos(ax)+(1-c) \cos(bx)=0$

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.

• 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) – ExtremeRaider May 19 at 4:51
• 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 $c$. – ExtremeRaider May 19 at 4:58