We have to find the general solution to:
[tex]3\cot \theta=-\sqrt[]{3}[/tex]
We can start rearranging the equation:
[tex]\begin{gathered} \cot \theta=-\frac{\sqrt[]{3}}{3} \\ \frac{1}{\tan\theta}=-\frac{\sqrt[]{3}}{3} \\ \frac{1}{\tan\theta}=-\frac{1}{\sqrt[]{3}} \\ \tan \theta=-\sqrt[]{3} \end{gathered}[/tex]
We will have one solution per cycle of length π.
We calculate the solution for the first period as:
[tex]\begin{gathered} \tan \theta=-\sqrt[]{3} \\ \theta=\arctan (-\sqrt[]{3}) \\ \theta=\frac{2\pi}{3} \end{gathered}[/tex]
We then can generalize for the other periods as:
[tex]\theta=\frac{2\pi}{3}+n\pi[/tex]
Answer: θ = 2π/3 + πn
