8 sec2 x + 4 tan2 x − 12 = 0

$$\begin{array}{rlll} 8sec^2x+4tan^2x-12&=&0\\\\ \frac{8}{cos^2x}+4(\frac{1}{cos^2x}-1)-12&=&0\\\\ \frac{8}{cos^2x}+\frac{4}{cos^2x}-4-12&=&0\\\\ \frac{12}{cos^2x}-16&=&0\\\\ \frac{12}{cos^2x}&=&16\\\\ \frac{3}{cos^2x}&=&4\\\\ \frac{3}{4}&=&cos^2x\\\\ cos^2x&=&\frac{3}{4}\\\\ cos\;x&=&\frac{\pm \sqrt3}{2}\\\\ x&=&n\pi\pm \frac{\pi}{6}\qquad \mbox{where}\;\;n \in Z\\\\ \end{array}$$

(2 sin² x − 1)(3 tan² x − 1) = 0