## Trigonometry (11th Edition) Clone

$\left\{\dfrac{\pi}{8}, \dfrac{3\pi}{8}, \dfrac{5\pi}{8}, \dfrac{7\pi}{8}, \dfrac{9\pi}{8}, \dfrac{11\pi}{8}, \dfrac{13\pi}{8}, \dfrac{15\pi}{8}\right\}$
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: $\left\{\dfrac{\pi}{8}, \dfrac{3\pi}{8}, \dfrac{5\pi}{8}, \dfrac{7\pi}{8}, \dfrac{9\pi}{8}, \dfrac{11\pi}{8}, \dfrac{13\pi}{8}, \dfrac{15\pi}{8}\right\}$