How to see the symmetry in this trigonometric equation Consider the equation $2\cos^2x-\cos x-1=0$. We can factor the LHS to obtain: $$(2\cos x + 1)(\cos x-1)=0,$$ leading to three solutions in the interval $[0,2\pi)$, namely $x=0, \frac{2\pi}{3}, \frac{4\pi}{3}$. If we want all solutions over $\Bbb R$, then we can add any multiple of $2\pi$ to each of these solutions.
However, all solutions over $\Bbb R$ are more efficiently expressed as simply the integer multiples of $\frac{2\pi}{3}$. That kind of solution is what one would expect from a problem that started out with something like $\cos (3x)=1$. However, if we expand $\cos(3x)$ using sum formulas, that equation leads us to a different polynomial in $\cos(x)$, namely: $$(2\cos x+1)^2(\cos x-1)=0.$$
So, it's clear that both polynomials have the same roots, and that "explains" why the solution sets are the same. Great.
My question: Is there a reasonable way to recognize the symmetry in the original equation, and transform it into an equation in $\cos(3x)$, other than just knowing ahead of time how it's going to work out?

$2\cos^2x-\cos x-1=(\cos{2x}+1)-\cos x-1=\cos 2x-\cos x$
$\cos 2x -\cos x=-2\sin{\frac{3x}{2}}\sin{\frac{x}{2}}$
Is it OK at this point?
You can play more with the trigonometric identities to see sometimes if it does turn out well. 
