## Task A-3.1. (4 points)

Determine all $x \in[0,2 \pi]$ for which $3 \sin x$ and $\frac{4}{7} \cos (2 x)+\frac{5}{3} \sin x$ are integers.
## Solution.

Let $n=3 \sin x \in \mathbb{Z}$. Then $-3 \leqslant n \leqslant 3$, so $n \in\{-3,-2,-1,0,1,2,3\}$. (1 point) The expression $\frac{4}{7} \cos (2 x)+\frac{5}{3} \sin x$ is then equal to:

$$
\begin{aligned}
& \frac{4}{7}\left(\cos ^{2} x-\sin ^{2} x\right)+\frac{5}{3} \sin x=\frac{4}{7}\left(1-2 \sin ^{2} x\right)+\frac{5}{3} \sin x \\
& =\frac{4}{7}\left(1-2 \cdot \frac{n^{2}}{9}\right)+\frac{5}{3} \cdot \frac{n}{3}=\frac{-8 n^{2}+35 n+36}{63}
\end{aligned}
$$

By substituting the numbers from the set $\{-3,-2,-1,0,1,2,3\}$ into the last expression, we find that it is an integer only for $n=1$.

(1 point)

From this, we conclude that the conditions of the problem are satisfied by all real numbers $x$ for which $3 \sin x=1$, or $\sin x=\frac{1}{3}$. In the given interval $[0,2 \pi]$, we have two solutions: $x_{1}=\arcsin \frac{1}{3}$ and $x_{2}=\pi-\arcsin \frac{1}{3}$.

(1 point)