Answer
$\left\{\dfrac{\pi}{8}, \dfrac{3\pi}{8}, \dfrac{5\pi}{8}, \dfrac{7\pi}{8}, \dfrac{9\pi}{8}, \dfrac{11\pi}{8}, \dfrac{13\pi}{8}, \dfrac{15\pi}{8} \right\}$
Work Step by Step
Take the square root of both sides to obtain:
$\sqrt{\sec^2{(2x)}}=\pm\sqrt2
\\\sec{(2x)}=\pm\sqrt2$
This means $\sec{(2x)}=-\sqrt2$ or $\sec{(2x)}=\sqrt2$
RECALL:
$\sec{x} = \frac{1}{\cos{x}}$
Thus,
$\sec{(2x)}=-\sqrt2 \longrightarrow \frac{1}{\cos{(2x)}}=-\sqrt2$
and
$\sec{(2x)}=\sqrt2\longrightarrow \frac{1}{\cos{(2x)}}=\sqrt2$
Solve each equation to obtain:
\begin{array}{ccc}
&\frac{1}{\cos{(2x)}}=-\sqrt2 &\text{or} &\frac{1}{\cos{(2x)}} = \sqrt2
\\&1=-\sqrt2(\cos{(2x)}) &\text{or} &1=\sqrt2(\cos{(2x)})
\\&\frac{1}{-\sqrt2}=\cos{(2x)} &\text{or} &\frac{1}{\sqrt2} = \cos{(2x)}
\\&-\frac{\sqrt2}{2}=\cos{(2x)} &\text{or} &\frac{\sqrt2}{2}=\cos{(2x)}
\\&\cos^{-1}{\left(-\frac{\sqrt2}{2}\right)}=2x &\text{or} &\cos^{-1}{\left(\frac{\sqrt2}{2}\right)}=2x
\\&\dfrac{\cos^{-1}{\left(-\frac{\sqrt2}{2}\right)}}{2}=x &\text{or} &\dfrac{\cos^{-1}{\left(\frac{\sqrt2}{2}\right)}}{2}=x
\end{array}
Use a scientific calculator's inverse cosine function to obtain:
\begin{array}{ccc}
&\dfrac{\frac{3\pi}{4}}{2}=x &\text{or} &\dfrac{\frac{\pi}{4}}{2}=x
\\&\frac{3\pi}{8}=x &\text{or} &\frac{\pi}{8}=x
\end{array}
Note that for $x=\frac{5\pi}{8}, \frac{11\pi}{8}, \frac{13\pi}{8}$, $\cos{(2x)}=-\frac{\sqrt2}{2}$.
Note also that for $x=\frac{7\pi}{8}, \frac{9\pi}{8}, \frac{15\pi}{8}$, $\cos{(2x)}=\frac{\sqrt2}{2}$.
Therefore, the solution set of the given equation is:
$\left\{\frac{\pi}{8}, \frac{3\pi}{8}, \frac{5\pi}{8}, \frac{7\pi}{8}, \frac{9\pi}{8}, \frac{11\pi}{8}, \frac{13\pi}{8}, \frac{15\pi}{8} \right\}$
