How solve this equation $\sin x\cdot \sin20=2\sin(110-x) (\sin10)^2$ let 
$0<x<90$, and such
$$\sin x\cdot \sin20=2\sin{(110-x)}(\sin10)^2$$
find the $x$
my idea: since 
$$\sin x\cdot 2\sin10\cos10=2\sin(70+x)(\sin10)^2$$
so
$$\cot10=\dfrac{\sin(70+x)}{\sin x}$$
then How find it?
 A: $$\sin20^\circ=2\sin10^\circ\cos10^\circ$$
$$\implies\sin x\cos10^\circ=\sin(70^\circ+x)\sin10^\circ$$
$$\implies2\sin x\cos10^\circ=2\sin(70^\circ+x)\cos80^\circ$$
Using Werner’s Formula,
$$\sin(x-10^\circ)+\sin(x+10^\circ)=\sin(x+150^\circ)+\sin(x-10^\circ)$$
$$\implies x+10^\circ=180^\circ n+(-1)^n(x+150^\circ)$$ where $n$ is any integer
If $n$ is even $=2m$(say), $$x+10^\circ=180^\circ(2m)+(x+150^\circ)\iff 360^\circ  m+140^\circ=0\text{ (Is it possible?)}$$
If $n$ is odd, $=2m+1$(say), $$x+10^\circ=180^\circ(2m+1)-(x+150^\circ)$$
$$\iff x=90^\circ(2m+1)-70^\circ=\cdots$$
A: How about this (assuming degrees throughout):
$$\begin{align*}
\sin x \sin 20 &= 2 \sin (110 - x) \sin^2 10 \\
&= 2 \cos (20 - x) \sin^2 10 \\
&= 2 (\cos 20 \cos x + \sin 20 \sin x) \sin^2 10 \\
&= (\cos 20 \cos x + \sin 20 \sin x) (1 - \cos 20).
\end{align*}
$$
Then 
$$\begin{align*}
\tan x &= \frac{\cos 20 (1 - \cos 20)}{\sin 20 \cos 20} \\
&= \frac{1 - \cos 20}{\sin 20} \\
&= \tan 10,
\end{align*}
$$
so $x = 10^{\circ}$.
A: $2\sin(x)\cdot\sin10\cdot\cos10=2[\cos20\cdot\cos x+\sin20\cdot\sin x]\cdot(\sin10)^2$
$\dfrac{\sin x\cdot\cos10}{\sin10}=\cos(20-x)$
$\dfrac{\cos10}{\sin10}=\dfrac{\cos(20-x)}{\sin x}$
$\Longrightarrow x=10$
