## Precalculus (6th Edition) Blitzer

The solutions in the interval $[0,2\pi )$ are $\frac{5\pi }{9}$, $\frac{11\pi }{9}$, and $\frac{17\pi }{9}$.
We know that the period of the cotangent function is $\pi$. In the interval $(0,\,\,\pi ]$ the only value for which the tangent function is $-\sqrt{3}$ is $\frac{5\pi }{6}$. Therefore, all the solutions to $\cot \frac{3\theta }{2}=-\sqrt{3}$ are given by \begin{align} & \frac{3\theta }{2}=\frac{5\pi }{6}+n\pi \\ & \theta =\frac{5\pi }{9}+\frac{2n\pi }{3} \end{align} Where, n is any integer. The solutions in the interval $[0,2\pi )$ are obtained by letting $n=0$, $n=1$, and $n=2$. The equation is calculated by first taking $n$ as 0 and then as 1 and 2. It can be further simplified as follows: \begin{align} & \theta =\frac{5\pi }{9}+\frac{2n\pi }{3} \\ & =\frac{5\pi }{9}+\frac{2\times 0\times \pi }{3} \\ & =\frac{5\pi }{9}+0 \\ & =\frac{5\pi }{9} \end{align} \begin{align} & \theta =\frac{5\pi }{9}+\frac{2n\pi }{3} \\ & =\frac{5\pi }{9}+\frac{2\times 1\times \pi }{3} \\ & =\frac{5\pi }{9}+\frac{2\pi }{3} \\ & =\frac{5\pi +6\pi }{9} \end{align} $=\frac{11\pi }{9}$ \begin{align} & \theta =\frac{5\pi }{9}+\frac{2n\pi }{3} \\ & =\frac{5\pi }{9}+\frac{2\times 2\times \pi }{3} \\ & =\frac{5\pi }{9}+\frac{4\pi }{3} \\ & =\frac{5\pi +12\pi }{9} \end{align} $=\frac{17\pi }{9}$ Thus, the actual solutions in the interval $[0,2\pi )$ will be $\frac{5\pi }{9}$, $\frac{11\pi }{9}$, and $\frac{17\pi }{9}$.