Agakhanov N.K.

Find all angles $\alpha$ for which the set of numbers $\sin \alpha, \sin 2 \alpha, \sin 3 \alpha$ coincides with the set $\cos \alpha, \cos 2 \alpha, \cos 3 \alpha$.
$\alpha=\frac{\pi}{8}+\frac{\pi \cdot 2}{2}$.

First solution. From the coincidence of sets, it follows that

$$
\sin x+\sin 2 x+\sin 3 x=\cos x+\cos 2 x+\cos 3 x, \text { i.e., } \sin 2 x(1+2 \cos x)=\cos 2 x(1+2 \cos x)
$$

a. $1+2 \cos x=0 \Rightarrow \cos x=-\frac{1}{2} \Rightarrow x= \pm \frac{2 \pi}{3}+2 \pi n \Rightarrow \sin 3 x=0$, but 0 does not belong to $\{\cos x, \cos 2 x, \cos 3 x\}$.
b. $\sin 2 x=\cos 2 x \Rightarrow \operatorname{tg} 2 x=1, x=\frac{\pi}{8}+\frac{\pi m}{2}$. The sets coincide, since $3 x+x=\frac{\pi}{2}+2 \pi n \Rightarrow \cos 3 x=\sin x, \sin 3 x=\cos x$.

Second solution. Add three unit vectors forming angles $\alpha, 2 \alpha, 3 \alpha$ with the $O x$ axis, respectively. By the condition, the coordinates of the resulting vector along $x$ and $y$ are equal, since they are the sums of the same three numbers. Therefore, this vector, if it is not zero, is directed along the line forming an angle $\pi / 4$ with the $O x$ axis. But the direction of the sum of three vectors coincides with the direction of the vector forming an angle $2 \alpha$, since the other two are symmetric relative to it.

Thus, $2 \alpha=\frac{\pi}{4}+\pi k$, from which $\alpha=\frac{\pi}{8}+\frac{\pi k}{2}$. Direct verification shows that all angles of the specified form fit.

It remains to consider the case when the sum of three unit vectors is zero. It is easy to see that this is possible only if the angles between them are $\frac{2 \pi}{3}$, from which $\alpha=\frac{2 \pi}{3}+\pi k$ or $\alpha=\frac{4 \pi}{3}+\pi k$. It is easy to see that then $\sin 3 \alpha=0, \cos \alpha \neq 0, \cos 2 \alpha \neq 0$ and $\cos 3 \alpha \neq 0$.

Problem