$9 \cdot 23$ Find all real numbers $\alpha$ such that
$\cos \alpha, \cos 2 \alpha, \cos 4 \alpha, \cdots, \cos 2^{n} \alpha, \cdots$

are all negative.
[Solution] Let $\alpha$ satisfy the requirements of the problem, then from $\cos 4 \alpha<0$ and $\cos 2 \alpha<0$ we get
$$
-\frac{\sqrt{2}}{2}<\cos 2 \alpha<0 \text {. }
$$

Furthermore, by $\cos \alpha<0$ we know $\cos \alpha<-\frac{\sqrt{2-\sqrt{2}}}{2}<-\frac{1}{4}$. This immediately yields
$$
\cos 2^{n} \alpha<-\frac{1}{4}, n=0,1,2, \cdots
$$

Therefore, we have $\left|\cos 2^{n} \alpha-\frac{1}{2}\right| \geqslant \frac{3}{4}, n=0,1,2, \cdots$
Since $\left|\cos ^{2 n+1} \alpha+\frac{1}{2}\right|=2\left|\cos ^{2} 2^{n} \alpha-\frac{1}{4}\right|$
$$
\begin{array}{l}
=2\left|\cos 2^{n} \alpha-\frac{1}{2}\right| \cdot\left|\cos 2^{n} \alpha+\frac{1}{2}\right| \\
\geqslant \frac{3}{2}\left|\cos 2^{n} \alpha+\frac{1}{2}\right|,
\end{array}
$$

Thus, $\left|\cos \alpha+\frac{1}{2}\right| \leqslant\left(\frac{2}{3}\right)^{n}, n=1,2,3, \cdots$
Hence, $\cos \alpha=-\frac{1}{2}$, i.e., $\alpha= \pm \frac{2 \pi}{3}+2 k \pi, k \in \mathbb{Z}$. On the other hand, when $\alpha= \pm \frac{2 \pi}{3}+$ $2 k \pi, k \in \mathbb{Z}$, it is clear that $\cos 2^{n} \alpha=-\frac{1}{2}, n=0,1,2, \cdots$