Let $P$ be a point inside the equilateral triangle $ABC$ such that $6\angle PBC = 3\angle PAC = 2\angle PCA$. Find the measure of the angle $\angle PBC$ .
Set $x=\angle PBC$. $0^\circ<\angle PCB=60^\circ-3x\implies 0^\circ< x<20^\circ$. By Ceva's
$$1=\frac{\sin(x)}{\sin(60^\circ-x)}\cdot\frac{\sin(3x)}{\sin(60^\circ-3x)}\cdot\frac{\sin(60^\circ-2x)}{\sin(2x)}$$Now set $t=\cos(2x)$. Of course $t\in (0,1)$.
Our equation is after some transformations (use identity $4\sin a\sin b\sin c=\sum_{cyc}\sin (a+b-c)-\sin(a+b+c)$) as the following
$$t^2+3t-4t^3=\sqrt{3(1-t^2)}\cdot t\iff$$$$(1-t)(3+4t)=\sqrt{3(1-t^2)}\iff$$(equivalence comes from the domain of $t$)
$$|(1-t)(3+4t)|=\sqrt{3(1-t^2)}\iff$$$$(1-t)(3+4t)^2=3(1+t)\iff$$$$t=-\frac{1}{2}\vee t=-\frac{\sqrt3}{2}\vee t=\frac{\sqrt3}{2}$$Hence $$\cos (2x)=\frac{\sqrt3}{2}\iff x=15^\circ$$