## Trigonometry (11th Edition) Clone

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