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
 A: 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.
A: Using Euler's formula $e^{ix} = \cos x+i\sin x$ we can simplify this a bit:
\begin{align}
\sin 2x\sin x + \cos^2x &= \frac1{2i}\left(e^{2ix}-e^{-2ix}\right)\frac1{2i}\left(e^{ix}-e^{-ix} \right) + \frac14\left(e^{ix} + e^{-ix}\right)^2\\
&= \frac14\left(-e^{3ix} + e^{ix} +e^{-ix}-e^{-3ix} + e^{2ix} + 2 + e^{-2ix} \right)
\end{align}
and
\begin{align}
\sin 5x\sin 4x + \cos^24x &= \frac1{2i}\left(e^{5ix}-e^{-5ix}\right)\frac1{2i}\left(e^{4ix}-e^{-4ix}\right)+\frac14\left(e^{4ix}+e^{-4ix}\right)^2\\
&= \frac14\left(-e^{9ix} + e^{ix} +e^{-ix} -e^{-9ix} + e^{8ix} + 2 + e^{-8ix} \right).
\end{align}
Factoring and subtracting out common terms, we have
$$
-(e^{3ix} +e^{-3ix}) +e^{2ix}+e^{-2ix} = -(e^{9ix}+e^{-9ix}) +e^{8ix}+e^{-8ix}, 
$$
and hence
$$
\cos9x + \cos2x  = \cos8x +\cos3x.
$$
From the identity
$$
\cos\theta + \cos\varphi = 2\cos\left(\frac{\theta+\varphi}2\right)\cos\left(\frac{\theta-\varphi}2\right)
$$
this becomes
$$
\cos\left(\frac{11}2x\right)\left(\cos\left(\frac72x\right) - \cos\left(\frac52x\right) \right) = 0.
$$
From the identity
$$
\cos\theta - \cos\varphi = -2\sin\left(\frac{\theta+\varphi}2\right)\sin\left(\frac{\theta-\varphi}2\right)
$$
this becomes
$$
\cos\left(\frac{11}2x\right)\sin(3x)\sin\left(\frac12x\right)=0.
$$
Hence the solutions are
$$
x = \frac{2\pi\left(n+\frac12\right)}{11},\ n=-5,-4,-3,-2,-1,0,1,2,3,4
$$
and
$$
x = \frac{n\pi}3,\ n=-2,-1,0,1,2,3.
$$
A: 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!
A: You should be able to transform the equation into polynomial of $\cos(x)$.
For example, $\sin(2x)\sin(x)=2\sin^2(x)\cos(x)=2(1-\cos^2(x))\cos(x)$
