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
Add $1$ to both sides of the equation to obtain:
$\tan^2{(2x)}=1$
Take the square root of both sides to obtain:
$\sqrt{\tan^2{(2x)}}=\pm\sqrt1
\\\tan{(2x)}=\pm1$
This means $\tan{(2x)}=-1$ or $\tan{(2x)}=1$.
RECALL:
$\tan{\theta}=x \longrightarrow \theta = \tan^{-1}{x}$
Thus,
$\tan{(2x)}=-1 \longrightarrow 2x=\tan^{-1}{(-1)}$
$\tan{(2x)}=1 \longrightarrow 2x=\tan^{-1}{(1)}$
Note that
$\tan{(\frac{\pi}{4})}=1$
$\tan{(-\frac{\pi}{4})}=-1$
Hence,
$2x=\tan^{-1}{(-1)}
\\\longrightarrow 2x=-\frac{\pi}{4}
\\\longrightarrow x=-\frac{\pi}{8}$
and
$2x=\tan^{-1}{1}
\\\longrightarrow 2x=\frac{\pi}{4}
\\\longrightarrow x=\frac{\pi}{8}$
The period of the function $y=\tan{(2x)}$ is $\frac{\pi}{2}$.
This means that the value of the function repeats every interval of $\frac{\pi}{2}$.
Thus, if $\dfrac{\pi}{8}$ is a solution of the equation, then the following are also solutions of the equation:
$\dfrac{\pi}{8}+\dfrac{\pi}{2}=\dfrac{5\pi}{8}$
$\dfrac{5\pi}{8} + \dfrac{\pi}{2} = \dfrac{9\pi}{8}$
$\dfrac{9\pi}{8}+\dfrac{\pi}{2}=\dfrac{13\pi}{8}$
Similarly, if $\dfrac{-\pi}{8}$ is a solution of the equation, then the following are also solutions of the equation:
$\dfrac{-\pi}{8}+\dfrac{\pi}{2}=\dfrac{3\pi}{8}$
$\dfrac{3\pi}{8} + \dfrac{\pi}{2} = \dfrac{7\pi}{8}$
$\dfrac{7\pi}{8}+\dfrac{\pi}{2}=\dfrac{11\pi}{8}$
$\dfrac{11\pi}{8}+\dfrac{\pi}{2}=\dfrac{15\pi}{8}$
Therefore, the solution set of the given equation is:
$\left\{\dfrac{\pi}{8}, \dfrac{3\pi}{8}, \dfrac{5\pi}{8}, \dfrac{7\pi}{8}, \dfrac{9\pi}{8}, \dfrac{11\pi}{8}, \dfrac{13\pi}{8}, \dfrac{15\pi}{8}\right\}$
