Solve $\sin 84^\circ \sin(54^\circ-x)=\sin 126^\circ \sin x$.

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 $30^\circ$.