# domain of $\sqrt {\cos^{-1}(\cos x)-\lfloor x\rfloor}$

Here is my question where I got stucked.

The domain of $\sqrt{\cos^{-1}(\cos x)-\lfloor x\rfloor}$ where $\lfloor \cdot\rfloor$ denotes the greatest integer function (floor function).

$\arccos(\cos x) = x$ when $x \in [0, \pi]$, in this case $\arccos(\cos x) - \lfloor x \rfloor \geq 0$. When $x \notin [0,\pi]$, when $x$ is negative everything will be good. When it is too large, $\arccos(\cos x) - \lfloor x \rfloor < 0$. We will only consider $x \in (\pi, 2\pi]$ as second case. In this case, $\arccos(\cos x) = 2\pi - x$, you don't want this be less than $\lfloor x \rfloor$, which might be $3, 4, 5, 6$. Actually only $3$ is possible. Then the domain is $(-\infty, 2\pi -3]$.