2. Solve the equation $\cos 8 x=\frac{14}{3}(\cos 2 x-\sin 2 x)^{2}-1$. In your answer, specify the number equal to the sum of the roots of the equation that belong to the interval $\left[\frac{11 \pi}{2} ; \frac{13 \pi}{2}\right]$, rounding this number to two decimal places if necessary.
Solution. The original equation is equivalent to the equation

$$
\begin{aligned}
1-2 \sin ^{2} 4 x=\frac{14}{3}(1-\sin 4 x)-1 & \Longleftrightarrow 2 \sin ^{2} 4 x-\frac{14}{3} \sin 4 x+\frac{8}{3}=0 \Longleftrightarrow \\
& \Longleftrightarrow \sin 4 t=1 \Longleftrightarrow x=\frac{\pi}{8}+\frac{\pi k}{2}, k \in \mathbb{Z}
\end{aligned}
$$

In the given interval, the roots are $6 \pi+\frac{\pi}{8}$ and $6 \pi-\frac{3 \pi}{8}$. Their sum is $\frac{47 \pi}{4}=36.913 \ldots \approx$ $36.91$.

Answer: 36.91.