Trabalhando a expressão:\begin{matrix} (\tan^2{x} - 1)(1 - \cot^2{x} ) = \tan^2{x} + \cot^2{x} -2 = 4 \end{matrix}Com isso,\begin{matrix}\tan^2{x} + \cot^2{x} = 6&\Rightarrow& \dfrac{\sin^4{x} + \cos^4{x} }{\sin^2{x} \cos^2{x} } = 6&\Rightarrow& \sin^4{x} + \cos^4{x} = 6 \sin^2{x} \cos^2{x} \end{matrix} $\color{orangered}{\text{Obs:}}$ $$(\sin^2{x} + \cos^2{x})^2 = \sin^4{x} + \cos^4{x} + 2\sin^2{x} \cos^2{x} = 1$$Continuando,\begin{matrix} 6 \sin^2{x} \cos^2{x} = 1 - 2\sin^2{x} \cos^2{x} \\ 8 \sin^2{x} \cos^2{x} = 1 \end{matrix}$\color{orangered}{\text{Obs:}}$ $ 2 \sin{x} \cos{x} = \sin{2x}$\begin{matrix} \sin^2{2x} = \dfrac{1}{2} &\Rightarrow&\sin{2x} = \pm \dfrac{\sqrt{2}}{2} &\therefore& 2x = \dfrac{\pi}{4} + \dfrac{k\pi}{2} \end{matrix}Conclui-se então:\begin{matrix} \boxed{x = \{ \pi/8 + k\pi/4 \ , \ k \in \mathbb{Z}\}} \end{matrix}\begin{matrix}Letra \ (D) \end{matrix}