How solve this equation $\sin x\cdot \sin20=2\sin(110-x) (\sin10)^2$

Question

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?

Answer 1

$$\sin20^\circ=2\sin10^\circ\cos10^\circ$$

$$\implies\sin x\cos10^\circ=\sin(70^\circ+x)\sin10^\circ$$

$$\implies2\sin x\cos10^\circ=2\sin(70^\circ+x)\cos80^\circ$$

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), $$x+10^\circ=180^\circ(2m+1)-(x+150^\circ)$$

$$\iff x=90^\circ(2m+1)-70^\circ=\cdots$$

Answer 2

$2\sin(x)\cdot\sin10\cdot\cos10=2[\cos20\cdot\cos x+\sin20\cdot\sin x]\cdot(\sin10)^2$

$\dfrac{\sin x\cdot\cos10}{\sin10}=\cos(20-x)$

$\dfrac{\cos10}{\sin10}=\dfrac{\cos(20-x)}{\sin x}$

$\Longrightarrow x=10$

Answer 3

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 $x = 10^{\circ}$.