## Precalculus (6th Edition) Blitzer

The solutions in the interval $[0,2\pi )$ are $\frac{\pi }{18}$, $\frac{7\pi }{18}$, $\frac{13\pi }{18}$, $\frac{19\pi }{18}$, $\frac{25\pi }{18}$, and $\frac{31\pi }{18}$.
We know that the period of the tangent function is $\pi$. In the interval $(0,\,\,\pi ]$, the only value for which the tangent function $\frac{\sqrt{3}}{3}$ is $\frac{\pi }{6}$. Therefore, all the solutions to $\tan 3x=\frac{\sqrt{3}}{3}$ are given by: \begin{align} & 3x=\frac{\pi }{6}+n\pi \\ & x=\frac{\pi }{18}+\frac{n\pi }{3} \end{align} Where, n is any integer. And the solutions in the interval $[0,2\pi )$ are obtained by letting $n=0$, $n=1$, $n=2$, $n=3$, $n=4$, and $n=5$. And the equation is calculated by taking first $n$ as 0 and then as 1, 2, 3, 4, and 5. It can be further simplified as follows. \begin{align} & x=\frac{\pi }{18}+\frac{n\pi }{3} \\ & =\frac{\pi }{18}+\frac{0\times \pi }{3} \\ & =\frac{\pi }{18}+0 \\ & =\frac{\pi }{18} \end{align} \begin{align} & x=\frac{\pi }{18}+\frac{n\pi }{3} \\ & =\frac{\pi }{18}+\frac{1\times \pi }{3} \\ & =\frac{\pi }{18}+\frac{1\pi }{3} \\ & =\frac{\pi +6\pi }{18} \end{align} $=\frac{7\pi }{18}$ \begin{align} & x=\frac{\pi }{18}+\frac{n\pi }{3} \\ & =\frac{\pi }{18}+\frac{2\times \pi }{3} \\ & =\frac{\pi }{18}+\frac{2\pi }{3} \\ & =\frac{\pi +12\pi }{18} \end{align} $=\frac{13\pi }{18}$ \begin{align} & x=\frac{\pi }{18}+\frac{n\pi }{3} \\ & =\frac{\pi }{18}+\frac{3\times \pi }{3} \\ & =\frac{\pi }{18}+\frac{3\pi }{3} \\ & =\frac{\pi +18\pi }{18} \end{align} $=\frac{19\pi }{18}$ \begin{align} & x=\frac{\pi }{18}+\frac{n\pi }{3} \\ & =\frac{\pi }{18}+\frac{4\times \pi }{3} \\ & =\frac{\pi }{18}+\frac{4\pi }{3} \\ & =\frac{\pi +24\pi }{18} \end{align} $=\frac{25\pi }{18}$ \begin{align} & x=\frac{\pi }{18}+\frac{n\pi }{3} \\ & =\frac{\pi }{18}+\frac{5\times \pi }{3} \\ & =\frac{\pi }{18}+\frac{5\pi }{3} \\ & =\frac{\pi +30\pi }{18} \end{align} $=\frac{31\pi }{18}$