Bulgarian Mathematical Olympiad

Solution:
The equality is equivalent to
$$\sin (n-1) \alpha=\frac{(n-1) \sin 2 \alpha}{2}$$
When $n \geq 4$ setting $\alpha=\frac{\pi}{4}$ gives
$$\sin \left((n-1) \frac{\pi}{4}\right)=\frac{n-1}{2} \geq \frac{3}{2}$$
a contradiction.
When $n=1$ and $n=3$ the equality (1) is an identity and when $n=2$ we have $\sin \alpha=\frac{\sin (2 \alpha)}{2}$, which is not true for $\alpha=\frac{\pi}{4}$. Therefore the answer is $n=1$ and $n=3$.