7.1. Solve the equation

$$
3 \cos \frac{4 \pi x}{5}+\cos \frac{12 \pi x}{5}=2 \cos \frac{4 \pi x}{5}\left(3+\operatorname{tg}^{2} \frac{\pi x}{5}-2 \operatorname{tg} \frac{\pi x}{5}\right)
$$

In the answer, write the sum of its roots on the interval $[-11 ; 19]$.
Answer: 112.5.

Solution. Let $\frac{\pi x}{5}=z$. Then, using the formula for the cosine of a triple angle, we get

$$
\begin{gathered}
3 \cos 4 z+4 \cos ^{3} 4 z-3 \cos 4 z=2 \cos 4 z \cdot\left(3+\operatorname{tg}^{2} z-2 \operatorname{tg} z\right) \\
\left(2 \cos ^{2} 4 z-3-\operatorname{tg}^{2} z+2 \operatorname{tg} z\right) \cos 4 z=0 \\
{\left[\begin{array}{l}
\cos 4 z=0 \\
2 \cos ^{2} 4 z=\operatorname{tg}^{2} z-2 \operatorname{tg} z+3
\end{array}\right.} \\
{\left[\begin{array}{l}
\cos 4 z=0, \\
2\left(\cos ^{2} 4 z-1\right)=(\operatorname{tg} z-1)^{2}
\end{array}\right.}
\end{gathered}
$$

Since in the second equation of the system, the left side is non-positive, and the right side is non-negative, we get from the first equation $4 z=\frac{\pi}{2}+2 \pi k$ (i.e., $z=\frac{\pi}{8}+\frac{\pi k}{4}$), and from the second $z=\frac{\pi}{4}+\pi n$, $k, n \in \mathbb{Z}$. Therefore, $x=\frac{5}{8}+\frac{5 k}{4}$ or $x=\frac{5}{4}+5 n, k, n \in \mathbb{Z}$.

From the first series, 24 numbers fall into the interval $[-11 ; 19]$ for $k \in[-9 ; 14]$, their sum is

$$
\frac{5}{8} \cdot 24+\frac{5}{4} \cdot(-9-8-\ldots-1+0+1+\ldots+14)=15+\frac{5}{4} \cdot(10+11+\ldots+14)=15+75=90
$$

From the second series, 6 numbers fall into the interval $[-11 ; 19]$ for $k \in[-2 ; 3]$, their sum is

$$
\frac{5}{4} \cdot 6+5 \cdot(-2-1+0+1+2+3)=\frac{15}{2}+15=\frac{45}{2}
$$

Total sum: $90+\frac{45}{2}=112.5$.