2. Solve the equation $(\cos 2 x-3 \cos 4 x)^{2}=16+\cos ^{2} 5 x$.
Answer: $x=\frac{\pi}{2}+k \pi, k \in \mathrm{Z}$.

Solution. Note that for any $x$, the inequality $-4 \leq \cos 2 x-3 \cos 4 x \leq 4$ holds, which implies that the left side of the equation does not exceed 16. At the same time, the right side of the equation is no less than 16. Therefore, equality can only be achieved when both conditions $(\cos 2 x-3 \cos 4 x)^{2}=16$ and $16+\cos ^{2} 5 x=16$ are satisfied simultaneously, from which $|\cos 2 x-3 \cos 4 x|=4, \cos 5 x=0$.

From the second equation, we get $x=\frac{\pi}{10}+\frac{k \pi}{5}, k \in \mathrm{Z}$. By substituting into the first equation ${ }^{4}$, we verify that only $x=\frac{\pi}{2}+k \pi, k \in \mathrm{Z}$ fits.