If $x>1$, then: $$\log_2x+\log _x2+2\cos{\alpha}=(\sqrt{\log_2x}-\frac1{\sqrt{\log_2x}})^2+2+2\cos{\alpha}\geq0$$and equality occurs if and only if $x=2$ and $\alpha=(2k+1)\pi$ where $k$ is any integer. If $x<1$, then: $$\log_2x+\log _x2+2\cos{\alpha}=-(\sqrt{-\log_2x}-\frac1{\sqrt{-\log_2x}})^2-2+2\cos{\alpha}\leq0$$This is true for all $x<1$ and real $\alpha$. Therefore, the answer is: $$\begin{cases}0<x<1\\\alpha\in\mathbb R\end{cases}\text{ or }\begin{cases}x=2\\\alpha=(2k+1)\pi&\text{for any integer }k\end{cases}$$