Answer
$150^{\mathrm{o}},\quad 210^{\mathrm{o}}$
Work Step by Step
Since $\sec\theta$ is negative, $\theta$ must lie in quadrants II or III.
Since the absolute value of $\sec\theta$ is $\displaystyle \frac{2\sqrt{3}}{3}$,
browsing through the table: Function Values of Special Angles,
the reference angle, $\theta^{\prime}$ must be $30^{\mathrm{o}}$.
Now, from
$\left[\begin{array}{lllll}
Quadr.: & I & II & III & IV\\
\theta' & \theta & 180^{o}-\theta & \theta-180^{o} & 360^{o}-\theta
\end{array}\right]$
In quadrant II
$\theta'=180^{o}-\theta$
so
$\theta=180^{\mathrm{o}}-\theta^{\prime}=180^{\mathrm{o}}-30^{0}=150^{\mathrm{o}}$
In quadrant III
$\theta'=\theta-180^{o}$
so
$\theta=180^{0}+\theta^{\prime}=180^{\mathrm{o}}+30^{\mathrm{o}}=210^{\mathrm{o}}$.
