2. [5 points] Solve the equation $\cos 7 x+\cos 3 x-\sqrt{2} \cos 10 x=\sin 7 x+\sin 3 x$.
Answer: $x=\frac{\pi}{20}+\frac{\pi k}{5}, x=\frac{\pi}{12}+\frac{2 \pi k}{3}, x=\frac{\pi}{28}+\frac{2 \pi k}{7}, k \in \mathbb{Z}$.

Solution. The given equation is equivalent to the following:

$2 \cos 5 x \cos 2 x-2 \sin 5 x \cos 2 x=\sqrt{2} \cos 10 x \Leftrightarrow 2 \cos 2 x(\cos 5 x-\sin 5 x)=\sqrt{2}\left(\cos ^{2} 5 x-\sin ^{2} 5 x\right) \Leftrightarrow$

$$
\left[\begin{array}{l}
{ \operatorname { cos } 5 x - \operatorname { sin } 5 x = 0 , } \\
{ \operatorname { cos } 5 x + \operatorname { sin } 5 x = \sqrt { 2 } \operatorname { cos } 2 x }
\end{array} \Leftrightarrow \left[\begin{array}{l}
{ \operatorname { tan } 5 x = 1 } \\
{ \operatorname { cos } ( 5 x - \frac { \pi } { 4 } ) = \operatorname { cos } 2 x }
\end{array} \Leftrightarrow \left[\begin{array}{l}
5 x=\frac{\pi}{4}+\pi k \\
5 x-\frac{\pi}{4}=2 x+2 \pi k \\
5 x-\frac{\pi}{4}=-2 x+2 \pi k, k \in \mathbb{Z}
\end{array}\right.\right.\right.
$$

Finally, we get $x=\frac{\pi}{20}+\frac{\pi k}{5}, x=\frac{\pi}{12}+\frac{2 \pi k}{3}, x=\frac{\pi}{28}+\frac{2 \pi k}{7}, k \in \mathbb{Z}$.