How to solve $\sin 2x \sin x+(\cos x)^2 = \sin 5x \sin 4x+(\cos 4x)^2$? How to solve $\sin 2x \sin x+(\cos x)^2 = \sin 5x \sin 4x+(\cos 4x)^2$?
\begin{align*}
2\cos x(\sin x)^2+(\cos x)^2 & = \frac{1}{2}(\cos x- \cos 9x) +(\cos 4x)^2\\
4\cos x(1-(\cos x)^2)+2(\cos x)^2 & = \cos x + \cos 9x +(2(\cos 4x)^2-1)+1\\
\cos x(4(1-(\cos x)^2 + 2\cos x -1) & = \cos 9x + \cos 8x + 1
\end{align*}
I don't see any way to get rid of $8x$ and $9x$ as arguments to have same angles from there

As $5-4=2-1$
$$2(\sin x\sin2x-\sin5x\sin4x)=2(\cos^24x-\cos^2x)$$
Using http://mathworld.wolfram.com/WernerFormulas.html
$$\cos x-\cos3x-(\cos x-\cos9x)=-2(\sin^4x-\sin^2x)$$
As $\dfrac{9-3}2=4-1,$
using http://mathworld.wolfram.com/ProsthaphaeresisFormulas.html and 
Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $
$$2\sin\dfrac{9x-3x}2\cdot\sin\dfrac{9x+3x}2=2\sin(4x+x)\sin(4x-x)$$
I should leave it here for you!
