For $0 < \theta < 360^\circ,\,$ solve $\,\cos\theta = -\frac{\sqrt3}{2}$. For $0 < \theta < 360^\circ,\,$ solve $\cos\theta = -\dfrac{\sqrt3}{2}$.
I got 120 and 210 degrees. But this doesn't match the textbook's solutions. Did I do something wrong?
 A: Your first solution is incorrect. The terminal side of the angle is in the second or third quadrant (because of the negative argument), so the reference angle of $30^\circ$ is measured from each side of $180^\circ$. I.e. we have that our angle must be $180^\circ \pm30^\circ$.
So $$\cos^{-1}\left(-\frac{\sqrt 3}2\right) = \theta,\; \theta \in \{150^\circ, 210^\circ\}$$
A: Since a reference angle of $30^\circ$ is the right one for getting a cosine of $\frac{\sqrt{3}}{2}$, the options are $30^\circ$,$150^\circ$,$210^\circ$, and $330^\circ$. However, we need an angle that gives $-\frac{\sqrt{3}}{2}$ , so it must be in the second or third quadrant, where cosine can be negative. This narrows it down to $150^\circ$ and $210^\circ$.
A: The answer of $210^{\circ}$ is correct. However, $120^{\circ}$ is not. Note that on the unit circle, the coordinates of the point at $120^{\circ}$ are $(-\frac{1}{2}, -\frac{\sqrt3}{2})$. You're looking for $150^{\circ}$.
In general, if you're given $\textrm{cos }\theta = a$, the solutions will be symmetric about $0^{\circ}$ if $a$ is positive, and symmetric about $180^{\circ}$ if it is negative.
Similarly, for $\textrm{sin }\theta = a$, the solutions will be symmetric about $90^{\circ}$ if $a$ is positive and symmetric about $270^{\circ}$ if negative.
In this case, $210^{\circ} = 180^{\circ} + 30^{\circ}$, so the other solution will be $30^{\circ}$ in the other direction--that is, $150^{\circ}$.
