<s>[INST] 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.
 [/INST] 
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 $x=\angle DBC=\angle NDB=70$.
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