## Precalculus (6th Edition) Blitzer

The solutions in the interval $[0,2\pi )$ are $\frac{\pi }{9}$, $\frac{4\pi }{9}$, $\frac{7\pi }{9}$, $\frac{10\pi }{9}$, $\frac{13\pi }{9}$, and $\frac{16\pi }{9}$.
We know that the period of the tangent function is $\pi$. In the interval $(0,\,\,\pi ]$, the only value for which the tangent function equals $\sqrt{3}$ is $\frac{\pi }{3}$. All the solutions to $\tan 3x=\sqrt{3}$ are given by: \begin{align} & 3x=\frac{\pi }{3}+n\pi \\ & x=\frac{\pi }{9}+\frac{n\pi }{3} \end{align} Where n is any integer. 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 }{9}+\frac{n\pi }{3} \\ & =\frac{\pi }{9}+\frac{0\times \pi }{3} \\ & =\frac{\pi }{9}+0 \\ & =\frac{\pi }{9} \end{align} \begin{align} & x=\frac{\pi }{9}+\frac{n\pi }{3} \\ & =\frac{\pi }{9}+\frac{1\times \pi }{3} \\ & =\frac{\pi }{9}+\frac{1\pi }{3} \\ & =\frac{\pi +3\pi }{9} \end{align} $=\frac{4\pi }{9}$ \begin{align} & x=\frac{\pi }{9}+\frac{n\pi }{3} \\ & =\frac{\pi }{9}+\frac{2\times \pi }{3} \\ & =\frac{\pi }{9}+\frac{2\pi }{3} \\ & =\frac{\pi +6\pi }{9} \end{align} $=\frac{7\pi }{9}$ \begin{align} & x=\frac{\pi }{9}+\frac{n\pi }{3} \\ & =\frac{\pi }{9}+\frac{3\times \pi }{3} \\ & =\frac{\pi }{9}+\frac{3\pi }{3} \\ & =\frac{\pi +9\pi }{9} \end{align} $=\frac{10\pi }{9}$ \begin{align} & x=\frac{\pi }{9}+\frac{n\pi }{3} \\ & =\frac{\pi }{9}+\frac{4\times \pi }{3} \\ & =\frac{\pi }{9}+\frac{4\pi }{3} \\ & =\frac{\pi +12\pi }{9} \end{align} $=\frac{13\pi }{9}$ \begin{align} & x=\frac{\pi }{9}+\frac{n\pi }{3} \\ & =\frac{\pi }{9}+\frac{5\times \pi }{3} \\ & =\frac{\pi }{9}+\frac{5\pi }{3} \\ & =\frac{\pi +15\pi }{9} \end{align} $=\frac{16\pi }{9}$