2. [5 points] Solve the equation $\frac{\cos 8 x}{\cos 3 x+\sin 3 x}+\frac{\sin 8 x}{\cos 3 x-\sin 3 x}=\sqrt{2}$.
Answer: $x=\frac{\pi}{44}+\frac{2 \pi k}{11}, k \neq 11 p+4, k \in \mathbb{Z}, p \in \mathbb{Z}$.

Solution. By bringing the fractions on the left side of the equation to a common denominator, we obtain

$$
\frac{\cos 8 x \cos 3 x-\cos 8 x \sin 3 x+\cos 3 x \sin 8 x+\sin 3 x \sin 8 x}{\cos ^{2} 3 x-\sin ^{2} 3 x}=\sqrt{2} \Leftrightarrow \frac{\cos 5 x+\sin 5 x}{\cos 6 x}=\sqrt{2}
$$

The last equation is equivalent to the system $\left\{\begin{array}{l}\cos 5 x+\sin 5 x=\sqrt{2} \cos 6 x, \\ \cos 6 x \neq 0 .\end{array}\right.$ The equation of the system gives

$$
\cos \left(5 x-\frac{\pi}{4}\right)=\cos 6 x \Leftrightarrow\left[\begin{array} { l } 
{ 5 x - \frac { \pi } { 4 } = 6 x + 2 \pi k , } \\
{ 5 x - \frac { \pi } { 4 } = - 6 x + 2 \pi k , k \in \mathbb { Z } }
\end{array} \Leftrightarrow \left[\begin{array}{l}
x=-\frac{\pi}{4}+2 \pi k \\
x=\frac{\pi}{44}+\frac{2 \pi k}{11}, k \in \mathbb{Z}
\end{array}\right.\right.
$$

Now consider the condition $\cos 6 x \neq 0$.

If $x=-\frac{\pi}{4}+2 \pi k$, then $\cos 6 x=\cos \left(-\frac{3 \pi}{2}+12 \pi k\right)=0$, i.e., the condition $\cos 6 x \neq 0$ is violated.

If $x=\frac{\pi}{44}+\frac{2 \pi k}{11}$, then $\cos 6 x=\cos \left(\frac{3 \pi}{22}+\frac{12 \pi k}{11}\right)$. Find those integer values of $n$ and $k$ for which the equality $\frac{3 \pi}{22}+\frac{12 \pi k}{11}=\frac{\pi}{2}+\pi n$ holds. We get $12 k=4+11 n, k=n-\frac{n-4}{12}$. Since $k \in \mathbb{Z}$ and $n \in \mathbb{Z}$, it follows that $p=\frac{n-4}{12} \in \mathbb{Z}$. Thus, $n=12 p+4, k=11 p+4$. The obtained values of the variable $k$ must be excluded. Finally, we get $x=\frac{\pi}{44}+\frac{2 \pi k}{11}, k \neq 11 p+4, k \in \mathbb{Z}$, $p \in \mathbb{Z}$.