$\displaystyle\cos(2^{n}\alpha)$ gets mad！

Easy to see that $\alpha =2k\pi \pm \frac{2\pi}{3}$ $(k\in \mathbb{Z})$ will meet our requests.So the answer is yes.

If you are interested ^_^, the following is the proof of the uniqueness.

First,let's show that : for all nonnegative integer $n$ ,we have $\displaystyle|cos(2^{n}\alpha)+\frac{1}{2}|\ge (\frac{5}{3})^{n}|cos\alpha+\frac{1}{2}|$ .

Do induction on $n$ .The case $n=0$ is obvious. Assume that for $n-1$ the relation holds.

Then for $n$ , $\displaystyle|cos(2^{n}\alpha)+\frac{1}{2}| =2|cos^{2}(2^{n-1}\alpha) -\frac{1}{4}|$

$\displaystyle=2|cos(2^{n-1}\alpha)-\frac{1}{2}||cos(2^{n-1}\alpha)+\frac{1}{2}|$

$\displaystyle\ge 2(\frac{1}{3}+\frac{1}{2})(\frac{5}{3})^{n-1}|cos\alpha +\frac{1}{2}|$ ,the relation also holds.

Now,since $\displaystyle -1\le cos(2^{n}\alpha) < -\frac{1}{3}$ , we see $|cos(2^{n}\alpha)+\frac{1}{2}|\le 1$ .

Therefore, $0\le |cos\alpha+\frac{1}{2}|\le (\frac{3}{5})^{n}$ for all nonnegative integer $n$ .

But $\lim_{n\rightarrow +\infty} (\frac{3}{5})^{n} =0$ , which implies $cos\alpha = -\frac{1}{2}$ .

So $\alpha =2k\pi \pm \frac{2\pi}{3}$ $(k\in \mathbb{Z})$ are the all values $\alpha$ can take.