Solve the equation giving the exact solutions which lie in $\left[0, 2\pi \right)$: $\;$ $\;$ $\cot^2 x = 3 \csc x - 3$
$\cot^2 x = 3 \csc x - 3$
i.e. $\;$ $\dfrac{\cos^2 x}{\sin^2 x} = \dfrac{3}{\sin x} - 3$
i.e. $\;$ $\cos^2 x = 3 \sin x - 3 \sin^2 x$
i.e. $\;$ $1 - \sin^2 x = 3 \sin x - 3 \sin^2 x$
i.e. $\;$ $2 \sin^2 x - 3 \sin x + 1 = 0$
i.e. $\;$ $2 \sin^2 x - 2 \sin x - \sin x + 1 = 0$
i.e. $\;$ $2 \sin x \left(\sin x - 1\right) - 1 \left(\sin x - 1\right) = 0$
i.e. $\;$ $\left(2 \sin x - 1\right) \left(\sin x - 1\right) = 0$
i.e. $\;$ $\sin x = \dfrac{1}{2}$ $\;$ OR $\;$ $\sin x = 1$
i.e. $\;$ $\sin x = \sin \left(\dfrac{\pi}{6}\right)$ $\;$ OR $\;$ $\sin x = \sin \left(\dfrac{\pi}{2}\right)$
i.e. $\;$ $x = n \pi + \left(-1\right)^n \times \left(\dfrac{\pi}{6}\right), \;\;\; n \in Z$
OR $\;$ $x = m \pi + \left(-1\right)^m \times \left(\dfrac{\pi}{2}\right), \;\;\; m \in Z$
In the range $\left[0, 2 \pi \right)$,
when $n = 0$, $\;$ $x = \dfrac{\pi}{6}$
when $n = 1$, $\;$ $x = \pi - \dfrac{\pi}{6} = \dfrac{5 \pi}{6}$
when $m = 0$, $\;$ $x = \dfrac{\pi}{2}$
$\therefore \;$ For the given equation, the exact solutions which lie in $\left[0, 2\pi \right)$ are:
$x = \dfrac{\pi}{6}, \; \dfrac{5 \pi}{6}, \; \dfrac{\pi}{2}$