6. Let $\omega=\cos \frac{\pi}{5}+i \sin \frac{\pi}{5}$. Then the equation with roots $\omega, \omega^{3}, \omega^{7}, \omega^{9}$ is
A. $x^{4}+x^{3}+x^{2}+x+1=0$
B. $x^{4}-x^{3}+x^{2}-x+1=0$
C. $x^{4}-x^{3}-x^{2}+x+1=0$
D. $x^{4}+x^{3}+x^{2}-x-1=0$
$$
\begin{array}{l}
(x-\omega)\left(x-\omega^{3}\right)\left(x-\omega^{7}\right)\left(x-\omega^{9}\right)=\left[x^{2}-\left(\omega+\omega^{9}\right) x+\omega^{10}\right]\left[x^{2}-\left(\omega^{3}+\omega^{7}\right) x+\omega^{10}\right] \\
=\left(x^{2}-2 \cos \frac{\pi}{5} x+1\right)\left(x^{2}-2 \cos \frac{3 \pi}{5} x+1\right) .
\end{array}
$$

Thus, the constant term of the equation is 1, and the coefficient of the linear term is $-2\left(\cos \frac{\pi}{5}+\cos \frac{3 \pi}{5}\right)=-4 \cos \frac{\pi}{5} \cos \frac{2 \pi}{5}$ $=-\frac{4 \sin \frac{\pi}{5} \cos \frac{\pi}{5} \cos \frac{2 \pi}{5}}{\sin \frac{\pi}{5}}=-1$. Therefore, the answer is $B$.