IIT-JEE 2011 · Paper 1, Mathematics question 65

Rearrange. $\dfrac1{\sin\frac\pi n}-\dfrac1{\sin\frac{3\pi}n}=\dfrac1{\sin\frac{2\pi}n}$. With $\theta=\frac\pi n$, the left side is $\dfrac{2\cos2\theta\sin\theta}{\sin\theta\sin3\theta}$, so
$$2\cos2\theta\sin2\theta=\sin3\theta\iff\sin4\theta=\sin3\theta.$$

Solve. $4\theta=\pi-3\theta$, i.e. $\theta=\frac\pi7$: $n=7$.

Answer: 7