Solving $\sin (100^\circ-x) \sin 20^\circ =\sin (80^\circ-x)\sin 80^\circ$ 
Solve for $x$ such that
$$\sin (100^\circ-x) \sin 20^\circ =\sin (80^\circ-x)\sin 80^\circ$$

First, I use the co-function formula:
$$\sin 80^\circ = \cos 10^\circ \tag{1}$$
Also,
$$\sin 20^\circ = 2\sin 10^\circ \cos 10^\circ \tag{2}$$
From these, I got
$$\sin(100^\circ-x)\cdot 2\sin 10^\circ =\sin (80^\circ-x) \tag{3}$$
I thought to use
$$2\sin a \sin b =\cos(a-b)-\cos(a+b) \tag{4}$$
but I'm stuck.
Help me please.

$$\dfrac{\sin(100^\circ-x)}{\sin(80^\circ-x)}=\dfrac{\sin80^\circ}{\sin20^\circ}$$
Use Componendo et Dividendo,
$$\dfrac{\sin(100^\circ-x)-\sin(80^\circ-x)}{\sin(100^\circ-x)+\sin(80^\circ-x)}=\dfrac{\sin80^\circ-\sin20^\circ}{\sin80^\circ+\sin20^\circ}$$
Use Prosthaphaeresis Formulas
$$\dfrac{\tan10^\circ}{\cot x}=\dfrac{\tan30^\circ}{\tan50^\circ}$$
$$\iff\tan x=\tan30^\circ\tan40^\circ\tan80^\circ$$
Now using Proving a fact: $\tan(6^{\circ}) \tan(42^{\circ})= \tan(12^{\circ})\tan(24^{\circ})$,
$$\tan20^\circ\tan(60^\circ-20^\circ)\tan(60^\circ+20^\circ)=\tan(3\cdot20^\circ)$$
$$\implies\tan x=\cot20^\circ$$
The rest should be easy.
More generally we have reached at $$\tan y\tan(60^\circ-y)\tan(60^\circ+y)\tan(90^\circ-3y)=1$$
See also : 
$\cot(x+110^\circ)=\cot(x+60^\circ)\cot x\cot(x-60^\circ)$
Proving a fact: $\tan(6^{\circ}) \tan(42^{\circ})= \tan(12^{\circ})\tan(24^{\circ})$
How can I find the following product? $ \tan 20^\circ \cdot \tan 40^\circ \cdot \tan 80^\circ.$
Proving:$\tan(20^{\circ})\cdot \tan(30^{\circ}) \cdot \tan(40^{\circ})=\tan(10^{\circ})$
