## Precalculus (6th Edition) Blitzer

The two values of $\theta$ is $\frac{5\pi }{6}$ and $\frac{11\pi }{6}$.
The expression is $\tan \theta =\frac{-\sqrt{3}}{3}$. Here, the reference angle is $\frac{\pi }{6}$ and $\theta$ lies in quadrants II or IV. Consider: The angle $\theta$ lies in quadrant II. \begin{align} & \theta =\pi -\frac{\pi }{6} \\ & =\frac{6\pi -\pi }{3} \\ & =\frac{5\pi }{3} \end{align} Consider: The angle $\theta$ lies in quadrant IV. \begin{align} & \theta =2\pi -\frac{\pi }{6} \\ & =\frac{12\pi -\pi }{6} \\ & =\frac{11\pi }{6} \end{align}