Constructively solving a trig equation 
Solve the equation $$\frac{\sin(18°+x)}{\sin(x)}=\frac{\sin48°}{\sin18°}$$

If we use a computer we quickly note that $x=12°$, which can be easily proven:
$$\sin18°=2\sin48°\sin12°=\cos36°-\cos60°$$
$$\iff 2\sin18°=2\cos36°-1$$
$$\iff2\sin18°=1-4\sin^218°$$
The last equation is true because if $x=18°$ we have
$$5x=90°\implies 3x=90-2x$$
$$\implies 3\sin x-4\sin^3 x=\cos2x=1-2\sin^2x$$
$$\implies (1-\sin x)(4\sin^2x+2\sin x -1)=0$$
Since $\sin 18°\neq1$, we have the desired equality.
But how can we possibly solve it constructively?

I found that
$$\cot x =\frac{\sin 48^{\circ}-\sin 18^{\circ}\cos 18^\circ}{\sin^2 18^{\circ}}$$
Let $y=12^\circ$. 
Note that
$$\begin{aligned} \sin y(2\sin 4y-\sin 3y)-\cos y(1-\cos 3y) & =\cos 3y-\cos 5y +\cos 4y -\cos y \\ & = 1-2\sin^2\frac{3y}{2}-\frac12-2\sin\frac{5y}{2}\sin\frac{3y}2 \\ & = \frac12(1-2\sin\frac{3y}{2}-4\sin^2\frac{3y}{2})\\ & =0\end{aligned}$$
The last step is true because $\frac{3y}2=18^{\circ}$ and due to the relation derived by the original poster.
Hence
$$\frac{2\sin 4y-\sin 3y}{1-\cos 3y} =\cot y$$
Can you finish it?
