<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] 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 $x = 10^{\circ}$.
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