Example 10. Find the value of $\cos \frac{2 \pi}{7}+\cos -\frac{4 \pi}{7}+\cos \frac{6 \pi}{7}$.
Let $7 \theta=2 n \pi(n=1,2,3)$, then $4 \theta$ $=2 n \pi-3 \theta, \cos 4 \theta=\cos 3 \theta, 2 \cos ^{2} 2 \theta-1$ $=4 \cos ^{3} \theta-3 \cos \theta, 2\left(2 \cos ^{2} \theta-1\right)^{2}-1$ $=4 \cos ^{3} \theta-3 \cos \theta$. Simplifying, we get $8 \cos ^{4} \theta-4 \cos ^{3} \theta$ $-8 \cos ^{2} \theta+3 \cos \theta+1=0$.

It is known that $1, \cos \frac{2 \pi}{7}, \cos -\frac{4 \pi}{7}$, $\cos \frac{6 \pi}{7}$ are the four roots of the equation $8 x^{4}-4 x^{3}-8 x^{2}+3 x+1=0$.
By Vieta's formulas, we have $1+\cos \frac{2 \pi}{7}+\cos \frac{4 \pi}{7}$
$$
\begin{array}{l}
+\cos \frac{6 \pi}{7}=\frac{4}{8}=\frac{1}{2} \\
\text { Therefore, } \cos \frac{2 \pi}{7}+\cos -\frac{4 \pi}{7}+\cos \frac{6 \pi}{7} \\
=-\frac{1}{2} .
\end{array}
$$