2. [5 points] Solve the equation $\frac{\cos 14 x}{\cos 5 x-\sin 5 x}-\frac{\sin 14 x}{\cos 5 x+\sin 5 x}=-\sqrt{2}$.
Answer: $x=\frac{3 \pi}{76}+\frac{2 \pi k}{19}, k \neq 19 p+2, k \in \mathbb{Z}, p \in \mathbb{Z}$.

Solution. By bringing the fractions on the left side of the equation to a common denominator, we get $\frac{\cos 14 x \cos 5 x+\cos 14 x \sin 5 x-\cos 5 x \sin 14 x+\sin 5 x \sin 14 x}{\cos ^{2} 5 x-\sin ^{2} 5 x}=-\sqrt{2} \Leftrightarrow \frac{\cos 9 x-\sin 9 x}{\cos 10 x}=-\sqrt{2}$.

The last equation is equivalent to the system $\left\{\begin{array}{l}\cos 9 x-\sin 9 x=-\sqrt{2} \cos 10 x, \\ \cos 10 x \neq 0 .\end{array}\right.$ The equation of the system gives

$$
\cos \left(9 x+\frac{\pi}{4}\right)=-\cos 10 x \Leftrightarrow\left[\begin{array} { l } 
{ \frac { x - \frac { \pi } { 4 } } { 2 } = \pi k + \frac { \pi } { 2 } , } \\
{ \frac { 1 9 x + \frac { \pi } { 4 } } { 2 } = \pi k + \frac { \pi } { 2 } , k \in \mathbb { Z } }
\end{array} \Leftrightarrow \left[\begin{array}{l}
x=\frac{5 \pi}{4}+2 \pi k \\
x=\frac{3 \pi}{76}+\frac{2 \pi k}{19}, k \in \mathbb{Z}
\end{array}\right.\right.
$$

Now consider the condition $\cos 10 x \neq 0$.

If $x=\frac{5 \pi}{4}+2 \pi k$, then $\cos 10 x=\cos \left(\frac{25 \pi}{2}+20 \pi k\right)=0$, i.e., the condition $\cos 10 x \neq 0$ is violated.

If $x=\frac{3 \pi}{76}+\frac{2 \pi k}{19}$, then $\cos 10 x=\cos \left(\frac{15 \pi}{38}+\frac{20 \pi k}{19}\right)$. Find those integer values of $n$ and $k$ for which the equality $\frac{15 \pi}{38}+\frac{20 \pi k}{19}=\frac{\pi}{2}+\pi n$ holds. We get $20 k=2+19 n, k=n+\frac{2-n}{20}$. Since $k \in \mathbb{Z}$ and $n \in \mathbb{Z}$, it follows that $p=\frac{2-n}{20} \in \mathbb{Z}$. Thus, $n=20 p+2, k=19 p+2$. The obtained values of the variable $k$ must be excluded. Finally, we get $x=\frac{3 \pi}{76}+\frac{2 \pi k}{19}, k \neq 19 p+2, k \in \mathbb{Z}$, $p \in \mathbb{Z}$.