451. Find all values of $\alpha$, for each of which the sequence

$$
\cos \alpha, \cos 2 \alpha, \cos 4 \alpha, \cos 8 \alpha, \ldots, \cos 2^{n} \alpha, \ldots
$$
451. Answer: $\alpha= \pm \frac{2 \pi}{3}+2 k \pi, k \in \boldsymbol{Z}$. If $\alpha$ satisfies the condition, then $\cos \alpha \leqslant-\frac{1}{4}$ (otherwise $\cos 4 \alpha>0$). Similarly, for any natural $n: \cos 2^{n} \alpha \leqslant-\frac{1}{4}$, hence $\left|\cos 2^{n} \alpha-\frac{1}{2}\right| \geqslant \frac{3}{4}$.

But then

$\left|\cos 2^{n} \alpha+\frac{1}{2}\right|=2\left|\cos 2^{n-1} \alpha-\frac{1}{2}\right| \cdot\left|\cos 2^{n-1} \alpha+\frac{1}{2}\right| \geqslant$

$$
\geqslant \frac{8}{2}\left|\cos 2^{n-1} \alpha+\frac{1}{2}\right|
$$

Thus,

$\left|\cos \alpha+\frac{1}{2}\right| \leqslant \frac{2}{3}\left|\cos 2 \alpha+\frac{1}{2}\right| \leqslant \ldots$

$$
\ldots \leqslant\left(\frac{2}{3}\right)^{n}\left|\cos 2^{n} \alpha+\frac{1}{2}\right| \leqslant\left(\frac{2}{3}\right)^{n-1}
$$

for any $n$. It follows that $\cos \alpha=-1 / 2$.