How to solve this trigonometric equation 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?
 A: $\sin170^\circ=\sin(90+80)^\circ=\cos80^\circ(?)$
$\cos170^\circ=\cos(90+80)^\circ==-\sin80^\circ(?)$
$$\implies\frac{2\sin(20^{\circ})\sin(170^{\circ})}{1+2\sin(20^{\circ})\cos(170^{\circ})}=\dfrac{2\sin20^\circ\cos80^\circ}{1-2\sin20^\circ\sin80^\circ}$$
Now, $1-2\sin20^\circ\sin80^\circ=1-(\cos60^\circ+\cos80^\circ)=\dfrac{1-2\cos80^\circ}2$
As $1-2\cos2y=1-2(2\cos^2y-1)=-\dfrac{\cos3y}{\cos y}$
$\implies1-2\cos80^\circ=-\dfrac{\cos120^\circ}{\cos40^\circ}=\dfrac1{2\cos40^\circ}$
$\implies1-2\sin20^\circ\cos10^\circ=\dfrac1{4\cos40^\circ}$
$$F=8\sin20^\circ\cos80^\circ\cos40^\circ=4\cos40^\circ\cos80^\circ\cdot\dfrac{2\cos20^\circ\sin20^\circ}{\cos20^\circ}$$
$$=2\cos80^\circ\cdot\dfrac{2\sin40^\circ\cos40^\circ}{\cos20^\circ}=\dfrac{2\sin80^\circ\cos80^\circ}{\cos20^\circ}=\dfrac{\sin160^\circ}{\cos20^\circ}=\dfrac{\sin(180-20)^\circ}{\cos20^\circ}=?$$
A: 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 $$\frac{4 \sin (20 {}^{\circ})}{\csc (10 {}^{\circ})-2}$$
A: 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.
