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.
 A: We need to solve
$$\sin100^{\circ}\sin20^{\circ}\cos{x}-\cos100^{\circ}\sin20^{\circ}\sin{x}=\sin^280^{\circ}\cos{x}-\cos80^{\circ}\sin80^{\circ}\sin{x}$$ or since
$$\cos80^{\circ}\sin80^{\circ}-\cos100^{\circ}\sin20^{\circ}\neq0,$$
$$\tan{x}=\frac{\sin^280^{\circ}-\sin100^{\circ}\sin20^{\circ}}{\cos80^{\circ}\sin80^{\circ}-\cos100^{\circ}\sin20^{\circ}}.$$
But $$\frac{\sin^280^{\circ}-\sin100^{\circ}\sin20^{\circ}}{\cos80^{\circ}\sin80^{\circ}-\cos100^{\circ}\sin20^{\circ}}=\frac{1-\cos160^{\circ}-\cos80^{\circ}+\cos120^{\circ}}{\sin160^{\circ}-\sin120^{\circ}+\sin80^{\circ}}=$$
$$=\frac{\sin30^{\circ}+\cos20^{\circ}-\cos80^{\circ}}{\sin80^{\circ}-2\sin20^{\circ}\cos40^{\circ}}=\frac{\sin30^{\circ}+\sin50^{\circ}}{4\sin20^{\circ}\cos20^{\circ}\cos40^{\circ}-2\sin20^{\circ}\cos40^{\circ}}=$$
$$=\frac{2\sin40^{\circ}\cos10^{\circ}}{4\sin20^{\circ}\cos20^{\circ}\cos40^{\circ}-2\sin20^{\circ}\cos40^{\circ}}=\frac{4\sin20^{\circ}\cos20^{\circ}\cos10^{\circ}}{4\sin20^{\circ}\cos20^{\circ}\cos40^{\circ}-2\sin20^{\circ}\cos40^{\circ}}=$$
$$=\frac{2\cos20^{\circ}\cos10^{\circ}}{2\cos20^{\circ}\cos40^{\circ}-\cos40^{\circ}}=\frac{2\cos20^{\circ}\cos10^{\circ}}{\cos60^{\circ}+\cos20^{\circ}-\cos40^{\circ}}=\frac{2\cos20^{\circ}\cos10^{\circ}}{\sin30^{\circ}+\sin10^{\circ}}=$$
$$=\frac{2\cos20^{\circ}\cos10^{\circ}}{2\sin20^{\circ}\cos10^{\circ}}=\cot20^{\circ}=\tan70^{\circ}.$$
Id est, $$x=70^{\circ}+180^{\circ}k,$$ where $k\in\mathbb Z$.
A: 
A geometric solution:
First draw triangle $ABC$ where $\angle BAC=20$ and $AB=AC$. Easy to find out $\angle ABC=\angle ACB=80$.
Now add point $D$ on line $AC$ such that $\angle CBD = x$, so $\angle ABD=80-x$ and $\angle BDC = 100-x$.
${sin(80-x)\over sin20}={sin(100-x)\over sin80} \implies {AD\over BD}={BC\over BD}\implies AD=BC$.
Now draw $NM$ pass through $D$ and parallel to $BC$ where $N$ is on $AB$ and $NM=AB$. Easy to see $ANM$ is congruent to $ABC$. Now $AM=AC$ and $\angle MAC=60$ so $AMC$ is equilateral. So $\angle NMC=60-20=40$. Also since $NM=AM=MC$ so $NMC$ is isosceles and $\angle MNC=70$. Note by symmetry triangles $NDC$ and $DNB$ are congruent and therefore $\angle NDB=70$. Since $ND$ and $BC$ are parallel $x=\angle DBC=\angle NDB=70$.
A: $$\dfrac{\sin(100^\circ-x)}{\sin(80^\circ-x)}=\dfrac{\sin80^\circ}{\sin20^\circ}$$
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:$\tan(20^{\circ})\cdot \tan(30^{\circ}) \cdot \tan(40^{\circ})=\tan(10^{\circ})$
