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

I put this to Wolfram Alpha with $x=\cos\theta$ and got an output of
$$(2 x - 1) (2 x + 1) (x - 1) (x + 1) (32 x^5 - 16 x^4 - 32 x^3 + 12 x^2 + 6 x - 1)$$
It it returned roots of
$$x\in\{-0.841254,-0.415415,0.142315,0.654861,0.959493\}$$
for the last factor, so you need to solve a quintic to get all the roots it seems.  
EDIT: Oops, spoke too soon. Cheating from @Math1000's answer, we can see now that
$$\begin{align}\frac{\sin\frac{11y}2}{\sin\frac y2}&=-\left(32\cos^5(\pi-y)-16\cos^4(\pi-y)-32\cos^3(\pi-y)\right.\\
&\quad\left.+12\cos^2(\pi-y)+6\cos(\pi-y)-1\right)\end{align}$$
And then $11y/2=n\pi$ so the roots are
$$\theta\in\{\frac{\pi}3,-\frac{\pi}3,\frac{2\pi}3,-\frac{2\pi}3,0,\pi,\frac{9\pi}{11},-\frac{9\pi}{11},\frac{7\pi}{11},-\frac{7\pi}{11},\frac{5\pi}{11},-\frac{5\pi}{11},\frac{3\pi}{11},-\frac{3\pi}{11},\frac{\pi}{11},-\frac{\pi}{11}\}$$
EDIT: After a nice walk in the chill air, the most direct path becomes clear: starting from
$$\sin2x\sin x+\cos^2x=\sin5x\sin4x+\cos^24x$$
We can apply $\sin A\sin B=\frac12\left(\cos(A-B)-\cos(A+B)\right)$ and $\cos^2A=\frac12(\cos2A+1)$ to obtain
$$\frac12(\cos x-\cos3x)+\frac12(\cos2x+1)=\frac12(\cos x-\cos9x)+\frac12(\cos8x+1)$$
Rearrange to
$$\cos9x+\cos2x-\cos8x-\cos3x=0$$
Then apply
$$\cos A+\cos B=2\cos\left(\frac{A+B}2\right)\cos\left(\frac{A-B}2\right)$$
Arriving at
$$\begin{align}0&=2\cos\left(\frac{11x}2\right)\cos\left(\frac{7x}2\right)-2\cos\left(\frac{11x}2\right)\cos\left(\frac{5x}2\right)\\
&=2\cos\left(\frac{11x}2\right)\left(\cos\left(\frac{7x}2\right)-\cos\left(\frac{5x}2\right)\right)\end{align}$$
And finally we may avail ourselves of
$$\cos A-\cos B=-2\sin\left(\frac{A+B}2\right)\sin\left(\frac{A-B}2\right)$$
So that
$$-4\cos\left(\frac{11x}2\right)\sin(3x)\sin\left(\frac x2\right)=0$$
Our solutions are therefore
$$\frac{11x}2=\left(n+\frac12\right)\pi$$
for $-5\le n\le4$ and
$$3x=n\pi$$
for $-2\le n\le3$, $n\in\mathbb{Z}$. This accounts for all $16$ solutions $\pmod{2\pi}$ with much less effort. If someone were to have whispered "$11$" in your ear this solution would have become immediately apparent.
