Answer
$-\frac{\pi}{3}+k\pi$
Work Step by Step
$\tan \theta=-\sqrt{3}$
$\theta=\tan^{-1}(-\sqrt{3})$
$\theta=-\frac{\pi}{3}$
The only solution in the interval $(-\frac{\pi}{2}, \frac{\pi}{2})$ is $\theta=-\frac{\pi}{3}$. Remember that tangent has period $\pi$, and we can add any multiple of $\pi$ to that solution and the original equation would still hold.
So the solution is $\theta=-\frac{\pi}{3}+k\pi$.
