[Solution] Clearly, $f\left(\frac{\pi}{6}\right)=\left|\frac{\sqrt{3}}{2}+\frac{\alpha}{2}\right|, f\left(\frac{5}{6} \pi\right)=\left|-\frac{\sqrt{3}}{2}+\frac{\alpha}{2}\right|$, thus
$$
\begin{aligned}
\max f(x) & \geqslant \frac{1}{2}\left(\left|\frac{\sqrt{3}}{2}+\frac{\alpha}{2}\right|+\left|-\frac{\sqrt{3}}{2}+\frac{\alpha}{2}\right|\right) \\
& \geqslant \frac{1}{2}\left[\left(\frac{\sqrt{3}}{2}+\frac{\alpha}{2}\right)-\left(-\frac{\sqrt{3}}{2}+\frac{\alpha}{2}\right)\right]=\frac{\sqrt{3}}{2}
\end{aligned}
$$

Therefore, we have $\quad M \geqslant \frac{\sqrt{3}}{2}$.
On the other hand, let $\alpha=0, \beta=-\frac{1}{6}$, then
$$
\begin{array}{l}
f(x)=\left|\cos x-\frac{1}{6} \cos 3 x\right|=\left|\frac{3}{2} \cos x-\frac{2}{3} \cos ^{3} x\right|. \text { It is easy to see that } \\
\max _{x} f(x)=\max _{-1 \leqslant y \leqslant 1}|g(y)|=\max _{0 \leqslant y \leqslant 1} g(y),
\end{array}
$$

where $g(y)=\frac{3}{2} y-\frac{2}{3} y^{3}$. Since
$$
g(y)-g\left(\frac{\sqrt{3}}{2}\right)=\left(y-\frac{\sqrt{3}}{2}\right)\left[\frac{3}{2}-\frac{2}{3}\left(y^{2}+\frac{\sqrt{3}}{2} y+\frac{3}{4}\right)\right],
$$

when $0 \leqslant y \leqslant \frac{\sqrt{3}}{2}$, by $y^{2}+\frac{\sqrt{3}}{2} y+\frac{3}{4} \leqslant \frac{9}{4}$ we get
$$
g(y)-g\left(\frac{\sqrt{3}}{2}\right) \leqslant 0,
$$

and when $\frac{\sqrt{3}}{2} \leqslant y \leqslant 1$, by $y^{2}+\frac{\sqrt{3}}{2} y+\frac{3}{4} \geqslant \frac{9}{4}$ we get
$$
g(y)-g\left(\frac{\sqrt{3}}{2}\right) \leqslant 0 .
$$

Therefore, we have $\quad \max _{x} f(x)=\max _{0 \leq y \leq 1} g(y)=g\left(\frac{\sqrt{3}}{2}\right)=\frac{\sqrt{3}}{2}$.
Thus, we get $\quad M \leqslant \frac{\sqrt{3}}{2}$.
Combining (1) and (2), we conclude that $M=\frac{\sqrt{3}}{2}$.
9.57 Let the function $f(x)=|\cos x+\alpha \cos 2 x+\beta \cos 3 x|$, where $\alpha, \beta$ are real numbers. Find: $M=\min _{\alpha, \beta} \max _{x} f(x)$.