Example 1. Find the value of $\cos \frac{2 \pi}{7}+\cos \frac{4 \pi}{7}+\cos \frac{6 \pi}{7}$
Let $\cos \frac{\pi}{7}=x$, then the original expression $=2 x^{2}-1+3 x$
$$
-4 x^{3}-x=4 x^{3}+2 x^{2}-1 \text {. }
$$

Notice: $\cos \frac{4 \pi}{7}=-\cos \frac{3 \pi}{7}$,
i.e., 2
$$
\begin{array}{l}
\left(2 x^{2}-1\right)^{2}-1=3 x-4 x^{3}, \\
\therefore(x+1)\left(8 x^{3}-4 x+1\right)=0 .
\end{array}
$$

But $x \neq-1,8 x^{3}-4 x^{2}-4 x+1=0$,
$$
\begin{aligned}
\therefore \text { the original expression } & =-\frac{1}{2}\left(8 x^{3}-4 x^{2}-4 x\right)-1 \\
& =-\frac{1}{2} .
\end{aligned}
$$
(Turn to page 41)