3. Find all values of $\alpha$ such that all terms in the sequence $\cos \alpha$, $\cos 2 \alpha, \cos 4 \alpha, \cos 8 \alpha, \cdots$ are negative.
The solution for the value of $\alpha$ is $\pm \frac{2 \pi}{3}+2 \pi m, m \in Z$. In fact, if $\alpha$ satisfies the conditions of the problem, then $\cos \alpha \leqslant -\frac{1}{4}$ (otherwise $-\frac{1}{4} < 0$). For the same reason, for any $n \in N$, we have $\cos 2^{2} \alpha \leqslant -\frac{1}{4}$, thus $\cos 2^{2} \alpha - \frac{1}{2} \geqslant \frac{3}{4}$. This leads to the conclusion.
$$
\operatorname{cose} x + \frac{1}{2} = 2\left(\cos x - \frac{1}{2}\right)\left(\cos x + \frac{1}{2}\right).
$$

We get
$$
\begin{array}{l}
\cos \alpha + \frac{1}{2} \leqslant \frac{2}{3} \cos 2 \alpha + \frac{1}{2} \leqslant \left(\frac{2}{3}\right)^{2} \\
\cdot \cos 4 \alpha + \frac{1}{2} \leqslant \cdots \leqslant \left(\frac{2}{3}\right)^{n} \cos 2 \pi + \frac{1}{2} \\
\leqslant \left(\frac{2}{3}\right)^{n}, n \in N.
\end{array}
$$

This indicates that $\cos \alpha + \frac{1}{2}$ can be smaller than any positive number, i.e., $\cos \alpha + \frac{1}{2} = 0$. For the specified $\alpha = \pm \frac{2 \pi}{3} + 2 \pi m$, we have
$$
0 > -\frac{1}{2} = \cos \alpha = \cos 2 \alpha = \cos 4 \alpha =
$$
$\cos 8 \alpha = \cdots$