Answer
$\left[0, \frac{\pi}{4}\right) \cup \left(\frac{3\pi}{4}, \frac{5\pi}{4}\right) \cup \left(\frac{7\pi}{4}, 2\pi\right]$
Work Step by Step
The tangent function is increasing on the intervals $\left(\frac{(2k-1)\pi}{2},\frac{(2k+1)\pi}{2}\right)$, where $k$ is integer.
In $[0,2\pi]$ the trangent function increases in the intervals $\left[0,\frac{\pi}{2}\right)$, $\left(\frac{\pi}{2},\frac{3\pi}{2}\right)$ and $\left(\frac{3\pi}{2},2\pi\right]$
$\tan 0 = 0$
$\tan \frac{\pi}{4} = 1$
So the interval $\left[0, \frac{\pi}{4}\right)$ is included
$\tan \frac{3\pi}{4} = -1$
$\tan \frac{5\pi}{4} = 1$
So the interval $\left(\frac{3\pi}{4}, \frac{5\pi}{4}\right)$ is included
$\tan \frac{7\pi}{4} = -1$
$\tan 2\pi = 0$
So the interval $\left(\frac{7\pi}{4}, 2\pi\right]$ is included
Finally we get that the values of $x$ on the interval $[0, 2\pi]$ that satisfy $ -1 < \tan x < 1 $ are:
$\left[0, \frac{\pi}{4}\right) \cup \left(\frac{3\pi}{4}, \frac{5\pi}{4}\right) \cup \left(\frac{7\pi}{4}, 2\pi\right]$
