Answer
$120^{\mathrm{o}}, \qquad 240^{\mathrm{o}}$
Work Step by Step
Since $\cos\theta$ is negative,
$\theta$ must lie in quadrants II or III.
Since the absolute value of $\cos\theta$ is $\displaystyle \frac{1}{2}$,
browsing through the table: Function Values of Special Angles,
the reference angle, $\theta^{\prime}$ must be $60^{\mathrm{o}}$.
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
$\theta=180^{0}-\theta^{\prime}=180^{0}-60^{0}=120^{\mathrm{o}}$,
and
In quadrant III ,
$\theta'=\theta-180^{o}$
so
$\theta= 180^{0}+\theta^{\prime}=180^{0}+60^{0}=240^{\mathrm{o}}$.
