ANSWER:
[tex]x=\frac{\pi}{6},\frac{11\pi}{6},\frac{5\pi}{6},\frac{7\pi}{6}[/tex]
STEP-BY-STEP EXPLANATION:
We have the following equation:
[tex]3\cdot(\sec x)^2-4=0[/tex]
Solving for x (We use the substitution method):
[tex]\begin{gathered} u=\sec x \\ \text{ replacing} \\ 3u^2-4=0 \\ u^2=\frac{4}{3} \\ u=\pm\sqrt[]{\frac{4}{3}}=\pm\frac{2\sqrt[]{3}}{3} \\ \text{ replacing} \\ \sec x=\pm\frac{2\sqrt[]{3}}{3} \\ x=\sec ^{-1}(\pm\frac{2\sqrt[]{3}}{3}) \\ x=\frac{\pi}{6},\frac{11\pi}{6},\frac{5\pi}{6},\frac{7\pi}{6} \end{gathered}[/tex]
