How to solve this trigonometric equation

Question

Plese help to solve this equation: $$ \sin x=2\sin20^{\circ}\sin\left(170^{\circ}-x\right)$$ I tried to convert this equation: $$\sin x=2\sin20^{\circ}\left(\sin170^{\circ}\cos x-\cos170^{\circ}\sin x\right)$$ $$\sin x\left(1+2\sin20^{\circ}cos170^{\circ}\right)=2\sin20^{\circ}\sin170^{\circ}\cos x$$ $$\tan x\left(1+2\sin20^{\circ}cos170^{\circ}\right)=2\sin20^{\circ}\sin170^{\circ}$$ It will be $$\tan x=\frac{2\sin20^{\circ}\sin170^{\circ}}{1+2\sin20^{\circ}\cos170^{\circ}}$$
But how to solve it without calculator?

Answer 1

$\sin170^\circ=\sin(90+80)^\circ=\cos80^\circ(?)$

$\cos170^\circ=\cos(90+80)^\circ==-\sin80^\circ(?)$

$$\implies\frac{2\sin(20^{\circ})\sin(170^{\circ})}{1+2\sin(20^{\circ})\cos(170^{\circ})}=\dfrac{2\sin20^\circ\cos80^\circ}{1-2\sin20^\circ\sin80^\circ}$$

Now, $1-2\sin20^\circ\sin80^\circ=1-(\cos60^\circ+\cos80^\circ)=\dfrac{1-2\cos80^\circ}2$

As $1-2\cos2y=1-2(2\cos^2y-1)=-\dfrac{\cos3y}{\cos y}$

$\implies1-2\cos80^\circ=-\dfrac{\cos120^\circ}{\cos40^\circ}=\dfrac1{2\cos40^\circ}$

$\implies1-2\sin20^\circ\cos10^\circ=\dfrac1{4\cos40^\circ}$

$$F=8\sin20^\circ\cos80^\circ\cos40^\circ=4\cos40^\circ\cos80^\circ\cdot\dfrac{2\cos20^\circ\sin20^\circ}{\cos20^\circ}$$ $$=2\cos80^\circ\cdot\dfrac{2\sin40^\circ\cos40^\circ}{\cos20^\circ}=\dfrac{2\sin80^\circ\cos80^\circ}{\cos20^\circ}=\dfrac{\sin160^\circ}{\cos20^\circ}=\dfrac{\sin(180-20)^\circ}{\cos20^\circ}=?$$

Answer 2

you can divide the equation by $$1+2\sin(20^{\circ})\cos(170^{\circ})$$ and you will get $$\tan(x)=\frac{2\sin(20^{\circ})\sin(170^{\circ})}{1+2\sin(20^{\circ})\cos(170^{\circ})}$$ the right Hand side can be simplified to $$\frac{4 \sin (20 {}^{\circ})}{\csc (10 {}^{\circ})-2}$$

Answer 3

Notice that

$$\sin20°=2\sin10°\cos10°,$$ and $10°$ is supplementary to $170°$.

Then with an obvious notation,

$$\sin x=4sc(s\cos x+c\sin x)$$

$$\tan x=\frac{4cs^2}{1-4c^2s}.$$

Unfortunately, I don't see any simplification.