The graph of $2\cos(y)-\cos(3x)-\cos(x)=0$ (here's a link) suggests to me that this expression can be factored on the left, but I haven't uncovered how. Any trig masters out there who can see the factors? Or can give an explanation as to why such a factorization is impossible?
2026-04-09 02:18:44.1775701124
factoring a trigonometric expression
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I don't think the entire left can be factored, since it has the separate $\cos y$ term. However if one moves things over the equation becomes $$2\cos y=\cos(3x)+\cos(x)=-2\cos(x)\cdot(2\sin^2 x-1),$$ so that for example one could maybe pick out the $x$ values going with the $y$ for which $\cos y=0,$ if that's any help.