$\begin{aligned} \sin 2 \theta=\cos 3 \theta \\ \Rightarrow \sin 2 \theta=\sin \left(90^{\circ}-3 \theta\right) \\ 2 \theta=90^{\circ}-3 \theta \quad \therefore \sin \theta=\sin 18^{\circ}=\frac{\sqrt{5}-1}{4} \\ & \Rightarrow 5 \theta=90^{\circ} \\ & \Rightarrow \theta=18^{\circ} \\ & \Rightarrow\left(\frac{1}{q}\right)(p)=1 \end{aligned}$
$\Rightarrow \mathrm{p}=\mathrm{q}$