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?