My original question was,
Solve
$\cos 5x+\cos2x=\cos 3x$
I was able to stop at
$\cos 2x=0$ $\\ \\$ or $\\$ $\sin2x \sin x=\frac14$
Is there a method to solve the equations of the second type?
My approach to those solutions
$$\cos 5x+\cos2x=\cos 3x$$
$$\cos 3x-\cos 5x-\cos2x=0$$
$$2\sin4x \sin x-\cos2x=0$$
$$2\cdot2\sin 2x\cos2x\cdot\sin x-\cos2x=0$$
$$\cos2x(4\sin2x\sin x-1)=0$$
Best Answer
$\sin 2x=2\sin x\cos x$, so $2\sin^2 x\cos x=\frac14$, or $8\cos x(1-\cos^2 x)-1=0$.
This cubic in $\cos x$ doesn't factorise nicely, so there is no pretty solution. But it has two roots in $(0,1)$, so solutions do exist.