Answer
$$x=\frac{\pi }{4}, \qquad x=\frac{3\pi }{4}, \\ x=\frac{5\pi }{4}, \qquad x=\frac{7\pi }{4}$$
Work Step by Step
We solve as follows:
$$\sec ^2 x -2=0 \quad \Rightarrow \quad \frac{1}{\cos ^2 x}=2 \quad \Rightarrow \quad \cos x = \pm \frac{\sqrt {2}}{2} \quad \Rightarrow \quad x=\frac{\pi }{4} + 2n\pi, \quad \text{ or } \quad x=\frac{3\pi }{4} + 2n \pi, \quad \text{ or } \quad x= \frac{5\pi }{4}+2n \pi, \quad \text{ or } \quad x=\frac{7\pi }{4} + 2n\pi, \quad n \in \mathbb{Z}$$So, $x=\frac{\pi }{4}$, $x=\frac{3\pi }{4}$, $x= \frac{5\pi }{4}$, and $x=\frac{7\pi }{4}$ are the only solutions in the interval $[0, 2\pi ]$.
