Solve the equation giving the exact solutions which lie in $\left[0, 2\pi \right)$: $\;$ $\;$ $\tan^3 x = 3 \tan x$
$\tan^3 x = 3 \tan x$
i.e. $\;$ $\tan x \left(\tan^2 x - 3\right) = 0$
i.e. $\;$ $\tan x = 0$ $\;$ OR $\;$ $\tan^2 x - 3 = 0$
i.e. $\;$ $\tan x = 0$ $\;$ OR $\;$ $\tan x = \pm \sqrt{3}$
Now, $\tan x = 0$ $\implies$ $x = n \pi, \;\;\; n \in Z$
OR $\;$ $\tan x = \sqrt{3} = \tan \left(\dfrac{\pi}{3}\right)$ $\implies$ $x = m \pi + \dfrac{\pi}{3}, \;\;\; m \in Z$
OR $\;$ $\tan x = - \sqrt{3} = \tan \left(- \dfrac{\pi}{3}\right)$ $\implies$ $x = q \pi - \dfrac{\pi}{3}, \;\;\; q \in Z$
In the range $\left[0, 2 \pi \right)$,
when $n = 0$, $\;$ $x = 0$
when $n = 1$, $\;$ $x = \pi$
when $m = 0$, $\;$ $x = \dfrac{\pi}{3}$
when $m = 1$, $\;$ $x = \pi + \dfrac{\pi}{3} = \dfrac{4 \pi}{3}$
when $q = 0$, $\;$ $x = - \dfrac{\pi}{3}$
$\left[- \dfrac{\pi}{3} \text{ is the same as } 2 \pi - \dfrac{\pi}{3} = \dfrac{5 \pi}{3} \right]$
when $q = 1$, $\;$ $x = \pi - \dfrac{\pi}{3} = \dfrac{2 \pi}{3}$
$\therefore \;$ For the given equation, the exact solutions which lie in $\left[0, 2\pi \right)$ are:
$x = 0, \; \dfrac{\pi}{3}, \; \dfrac{2 \pi}{3}, \; \pi, \; \dfrac{4 \pi}{3}, \; \dfrac{5 \pi}{3}$