Solve $\sin 84^\circ \sin(54^\circ-x)=\sin 126^\circ \sin x$. 
Find $x$ in degrees, where $$\sin 84^\circ\cdot \sin(54^\circ-x)=\sin126^\circ\cdot \sin x\,.$$

I tried to use trigonometry identities to transform the product in sums, but I can't simplify moreover. I know the result is 30° since I solved it in a calculator, but there must be an algebraic way.
For context, this equation comes from solving this problem:
How can I solve this geometry problem without trigonometry?

Use
$$\cos36 = \frac {\sin108}{2\sin36}
= \frac {\sin36+2\sin36\cos72}{2\sin36}=\frac12+\cos72$$
to factorize the equation as follows
$$\begin{align}
& \sin 84\sin(54-x)-\sin 54 \sin x \\
& =\cos 6 \cos 36 \cos x - ( \cos 36+\cos 6 \sin 36 )\sin x \\
& =\frac12(\cos30+\cos 42 ) \cos x 
- \left(\frac12 + \cos 72 + \frac12(\sin 42+ \sin30 )\right)\sin x \\
& =\frac12\left(\frac{\sqrt3}2+\cos 42 \right) \cos x 
- \left(\frac34 +\cos (42+30) + \sin30\sin 42\right)\sin x \\
& =\frac12\left(\frac{\sqrt3}2+\cos 42 \right) \cos x 
- \left(\frac34 + \frac{\sqrt3}2\cos42 \right)\sin x \\
& =\frac12\left(\frac{\sqrt3}2+\cos 42 \right) (\cos x -\sqrt3 \sin x) = 0
\end{align}$$
Thus,
$$\tan x = \frac1{\sqrt3}$$
and the angle in the source problem is $30^\circ$.
