2. Solve the equation $\cos 2 x+\cos 6 x+2 \sin ^{2} x=1$.

In the answer, specify the number equal to the sum of the roots of the equation that belong to the interval $A$, rounding this number to two decimal places if necessary.

$$
A=\left[\frac{m \pi}{6} ; \frac{(m+1) \pi}{6}\right], m=5
$$
Answer: 2.88 (exact value: $\frac{11 \pi}{12}$ ).

Solution. Since $\cos 2 x+2 \sin ^{2} x=1$, the original equation is equivalent to $\cos 6 x=0$. From all the roots of the series $x=\frac{\pi}{12}+\frac{\pi n}{6}$, only one root $x=\frac{\pi}{12}+\frac{5 \pi}{6}=\frac{11 \pi}{12}$ falls within the interval $A=\left[\frac{5 \pi}{6} ; \pi\right]$.