3. Solve the equation

$$
\sin \left(x+3^{0} \cdot \frac{2 \pi}{7}\right)+\sin \left(x+3^{1} \cdot \frac{2 \pi}{7}\right)+\ldots+\sin \left(x+3^{5} \cdot \frac{2 \pi}{7}\right)=1
$$
Answer: $x=-\frac{\pi}{2}+2 \pi n, \quad n \in \mathbb{Z}$.

Solution. First, note that

$$
\sin x+\sin \left(x+\frac{2 \pi}{7}\right)+\ldots+\sin \left(x+6 \cdot \frac{2 \pi}{7}\right)=0
$$

Indeed, seven points on the unit circle of the form $x+i \cdot \frac{2 \pi}{7}(i=0,1, \ldots, 6)$ form the vertices of a regular heptagon, so the sum of the radius vectors with endpoints at these points is zero (since rotating the set of these vectors by an angle $\frac{2 \pi}{7}$ around the origin does not change the set, and therefore does not change the vector equal to their sum, which means it is zero), as is the sum of their ordinates, i.e., the corresponding sines.

Secondly, considering the equalities

$$
\begin{array}{ll}
\sin \left(x+3^{2} \cdot \frac{2 \pi}{7}\right)=\sin \left(x+2 \cdot \frac{2 \pi}{7}\right), & \sin \left(x+3^{3} \cdot \frac{2 \pi}{7}\right)=\sin \left(x+6 \cdot \frac{2 \pi}{7}\right), \\
\sin \left(x+3^{4} \cdot \frac{2 \pi}{7}\right)=\sin \left(x+4 \cdot \frac{2 \pi}{7}\right), & \sin \left(x+3^{5} \cdot \frac{2 \pi}{7}\right)=\sin \left(x+5 \cdot \frac{2 \pi}{7}\right),
\end{array}
$$

we obtain that the left side of the original equation is equal to

$$
\sin \left(x+3^{0} \cdot \frac{2 \pi}{7}\right)+\sin \left(x+3^{1} \cdot \frac{2 \pi}{7}\right)+\ldots+\sin \left(x+3^{5} \cdot \frac{2 \pi}{7}\right)=0-\sin x
$$

and the equation itself can be written as

$$
-\sin x=1 \Leftrightarrow x=-\frac{\pi}{2}+2 \pi n, \quad n \in \mathbb{Z}
$$