Solve the equation: $\;$ $\tan^3 x - 1 + \dfrac{1}{\cos^2 x} - 3 \cot \left(\dfrac{\pi}{2} - x\right) = 3$
Given equation: $\;\;\;$ $\tan^3 x - 1 + \dfrac{1}{\cos^2 x} - 3 \cot \left(\dfrac{\pi}{2} - x\right) = 3$
i.e. $\;$ $\tan^3 x + \left(-1 + \sec^2 x\right) - 3 \tan x - 3 = 0$
i.e. $\;$ $\tan^3 x + \tan^2 x - 3 \tan x - 3 = 0$
i.e. $\;$ $\tan^2 x \left(1 + \tan x\right) - 3 \left(1 + \tan x\right) = 0$
i.e. $\;$ $\left(1 + \tan x\right) \left(\tan^2 x - 3\right) = 0$
$\implies$ $1 + \tan x = 0$ $\;$ or $\;$ $\tan^2 x - 3 = 0$
Case 1:
$1 + \tan x = 0$ $\implies$ $\tan x = -1$
$\begin{aligned}
i.e. \; x & = n \pi + \tan^{-1} \left(-1\right) \\\\
& = n \pi - \tan^{-1} \left(1\right) \\\\
& = n \pi - \dfrac{\pi}{4}, \;\; n \in I
\end{aligned}$
Case 2:
$\tan^2 x - 3 = 0$ $\implies$ $\tan x = \pm \sqrt{3}$
When $\;$ $\tan x = \sqrt{3} = \tan \left(\dfrac{\pi}{3}\right)$
$\implies$ $x = m \pi + \dfrac{\pi}{3}, \;\; m \in I$
When $\;$ $\tan x = - \sqrt{3}$
$\begin{aligned}
i.e. \; x & = k \pi + \tan^{-1} \left(-\sqrt{3}\right) \\\\
& = k \pi - \tan^{-1} \left(\sqrt{3}\right) \\\\
& = k \pi - \dfrac{\pi}{3}, \;\; k \in I
\end{aligned}$