## Precalculus (6th Edition)

$x\displaystyle \in\{\frac{\pi}{18},\frac{7\pi}{18},\frac{13\pi}{18},\frac{19\pi}{18}, \frac{25\pi}{18},\frac{31\pi}{18}\}$
If $0 \leq x < 2\pi$ then $0 \leq 3x < 6\pi$ So, $3x\in[0,6\pi)$, meaning that if 3x (a solution) is any number from $[0,2\pi)$, then $(3x+2\pi)$ and $(3x+4\pi)$ are also solutions. Divide the equation with 3. $\displaystyle \tan 3x=\frac{\sqrt{3}}{3}$ In $[0,2\pi),\ \tan \displaystyle \frac{\pi}{6}=\frac{\sqrt{3}}{3}, \tan \displaystyle \frac{7\pi}{6}=\frac{\sqrt{3}}{3}$ so, $3x$ can be $\displaystyle \frac{\pi}{6} \qquad$or $\displaystyle \qquad \frac{7\pi}{6}.$ 3x can also be $\displaystyle \frac{\pi}{6}+2\pi=\frac{13\pi}{6} \quad$or $\displaystyle \quad \frac{7\pi}{6}+2\pi=\frac{19\pi}{6}$ And, also $\displaystyle \frac{\pi}{6}+4\pi=\frac{25\pi}{6} \quad$or $\displaystyle \quad \frac{7\pi}{6}+4\pi=\frac{31\pi}{6}$ So, $3x\displaystyle \in\{\frac{\pi}{6},\frac{7\pi}{6},\frac{13\pi}{6},\frac{19\pi}{6}, \frac{25\pi}{6},\frac{31\pi}{6}\}$, and, since $(3x)\div 3=x,$ after dividing each corresponding angle with 3, $x\displaystyle \in\{\frac{\pi}{18},\frac{7\pi}{18},\frac{13\pi}{18},\frac{19\pi}{18}, \frac{25\pi}{18},\frac{31\pi}{18}\}$