Solution. Multiply by $\sin \pi x$. All $x=n, n \in \mathbb{Z}$ need to be removed from the answer, as they are not roots of the original equation. Next, using the equality $8 \sin \pi x \cos \pi x \cos 2 \pi x \cos 4 \pi x=\sin 8 \pi x$, we find

$$
\begin{aligned}
\sin 8 \pi x \cdot \cos 2 \pi x & =\sin \pi x \cdot \cos 9 \pi x \Longleftrightarrow \\
\sin 10 \pi x+\sin 6 \pi x & =\sin 10 \pi x-\sin 8 \pi x \Longleftrightarrow \\
\sin 6 \pi x & =\sin (-8 \pi x)
\end{aligned}
$$

From which $x=k / 7$ or $x=1 / 2+l, k, l \in \mathbb{Z}$.

The roots of the equation in the interval $[0 ; 1]$ are: $1 / 7, 2 / 7, 3 / 7, 4 / 7, 5 / 7, 6 / 7$ and $1 / 2$.

Answer: 3.5.
8-1. Find all common points of the graphs

$$
y=8 \cos \pi x \cdot \cos ^{2} 2 \pi x \cdot \cos 4 \pi x \quad \text { and } \quad y=\cos 9 \pi x
$$

with abscissas belonging to the segment $x \in[0 ; 1]$. In your answer, specify the sum of the abscissas of the found points.