<s>[INST] 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?
 [/INST] $2\sin(x)\cdot\sin10\cdot\cos10=2[\cos20\cdot\cos x+\sin20\cdot\sin x]\cdot(\sin10)^2$
$\dfrac{\sin x\cdot\cos10}{\sin10}=\cos(20-x)$
$\dfrac{\cos10}{\sin10}=\dfrac{\cos(20-x)}{\sin x}$
$\Longrightarrow x=10$
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