Solve   \( 2tan^2x-secx-6=0\ \)   in the interval \( \quad - \pi \leqslant x \leqslant \pi\)

\begin{align} 0 & = 2tan^2x-secx-4 \\ 0 & = 2(sec^2-1)-secx-4 \\ 0 & = 2sec^2x -2 -secx-4\\ 0 & = 2sec^2x-secx-6 \\ 0 & = 2sec^2x - 4secx + 3secx-6 \\ 0 & =2secx(secx-2)+3(secx-2) \\ 0 & = (2secx+3)(secx-2)\\ \\ \implies secx & = -\frac{3}{2} \quad and \quad secx=2 \\ So \quad \frac{1}{cosx} & = -\frac{3}{2} \\ \implies\ cosx & = -\frac{2}{3}\\ And \quad \frac{1}{cosx} & = 2 \\ \implies\ cosx & = \frac{1}{2} \\ \\ \therefore\ \quad x & = 2.3^c\ and\ -2.3^c\,, \quad and \quad x= \frac{\pi}{3}\ and\ -\frac{\pi}{3} \\ \\ \end{align}