\(\theta = n\pi \)

\begin{align*}
\tan \left(\theta-\frac{\pi}{3}\right) &=-\sqrt{3} \\
\therefore \theta-\frac{\pi}{3} &=n \pi+\tan ^{-1}(-\sqrt{3}) \\
\theta-\frac{\pi}{3} &=n \pi-\frac{\pi}{3} \rightarrow \theta=n \pi
\end{align*}