$4 \cdot 85$ Let the equation $x^{4}+\frac{1}{3} \sin \alpha \cdot x^{2}+\frac{1}{200} \cos \frac{\pi}{3}=0$ have four roots that form an arithmetic sequence, where $\alpha$ is between 0 and $2 \pi$. Try to find this angle $\alpha$, and the values of these four roots.
[Solution] $x^{2}$ has two solutions $b$ and $c, x= \pm \sqrt{b}, \pm \sqrt{c}$, and since the four roots form an arithmetic sequence, and their arithmetic mean is zero, if the common difference is a positive number $2a$, then these four numbers should be $-3a, -a, a, 3a$. Here, $\pm a$ is the pair with the smaller absolute value among $\pm \sqrt{b}$ and $\pm \sqrt{c}$, and $\pm 3a$ is the other pair. From the product of the four roots,
$$
9a^{4} = \frac{1}{200} \cos \frac{\pi}{3} = \frac{1}{400},
$$

we get $\quad a = \pm \frac{1}{2 \sqrt{15}}, \pm \frac{i}{2 \sqrt{15}}$.
Comparing the relationship between roots and coefficients,
$$
\frac{1}{3} \sin \alpha = -(b+c) = -(9a^{2} + a^{2}) = -10a^{2} = \pm \frac{1}{6}.
$$

Thus, $\sin \alpha = \pm \frac{1}{2}$. Since $0 \leqslant \alpha < 2\pi$, we have
$$
\begin{array}{l}
\alpha = \frac{\pi}{6}, \frac{5\pi}{6} ; x = \pm \frac{i}{2 \sqrt{15}}, \pm \frac{3i}{2 \sqrt{15}}, \\
\alpha = \frac{7\pi}{6}, \frac{11\pi}{6} ; x = \pm \frac{1}{2 \sqrt{15}}, \pm \frac{3}{2 \sqrt{15}}.
\end{array}
$$