<s>[INST] Solve $\frac{\sin (10^\circ) \sin (30^\circ)}{\sin 40^\circ \sin (80^\circ-x^\circ)} = \frac{\sin 20^\circ}{\sin x}$ The context to this is trivial I think, I was solving a geometry problem using the trigonometric version of Ceva, I got here and I was stuck, I tried using the sum-difference, product to sums, sums to products identities but my attempts failed and was rather tedious.
 [/INST] You can use
\begin{equation}
\sin a\sin b\sin c=\frac{1}{4}(\sin(a+b-c)+\sin(a-b+c)+\sin(-a+b+c)-\sin(a+b+c))
\end{equation}
With this you get
\begin{equation}
\sin(40-x)+\sin(-20+x)+\sin(20+x)-\sin(40+x) \\
=
\sin(-20+x)+\sin(60-x)+\sin(100-x)-\sin(140-x)
\end{equation}
Then you can use $\sin(140 - x) = \sin(180 - 140 + x) = \sin(40+x)$ to reduce this to
\begin{equation}
\sin(40-x)+\sin(20+x) =\sin(60-x)+\sin(100-x) \\
\sin(70-(30+x))+\sin(-10+(30+x)) - \sin(90-(30+x)) - \sin(130-(30+x)) = 0 \\
\end{equation}
Now I use formula $\sin(a\pm(30+x))=\sin a\cos (30+x)\pm \cos a\sin (30+x)$.
\begin{equation}
\cos(30+x)(\sin 70 - \sin 10 - 1 -\sin130)+\sin(30+x)(-\cos 70 +\cos 10 + \cos 130) = 0 \\
-\cos(30+x)(\sin 250 + \sin 10 + 1 + \sin130)+\sin(30+x)(\cos 250 +\cos 10 + \cos 130) = 0
\end{equation}
Now we can notice the following
$(\cos 250 +\cos 10 + \cos 130) = \Re e^{i 10^\circ}(e^{i 0^\circ} + e^{i 120^\circ} + e^{i 240^\circ})$ and $(\sin 250 +\sin 10 + \sin 130) = \Im e^{i 10^\circ}(e^{i 0^\circ} + e^{i 120^\circ} + e^{i 240^\circ})$. But $e^{i 0^\circ} + e^{i 120^\circ} + e^{i 240^\circ} = 0$, which yields
\begin{equation}
-\cos(30+x) = 0
\end{equation}
which has two solutions $x=60$ and $x=-120$.
</s>