Question

I need help with this practice problem solving The subject is pre trigonometry

Answer 1

Answer:

$\theta=\frac{2\pi}{3}+\pi\cdot n$

Step-by-step explanation:

Find the general solution for:

$\begin{gathered} 3\cot \theta=-\sqrt[]{3} \\ \text{ Divide both sides by 3} \\ \cot \theta=-\frac{\sqrt[]{3}}{3} \end{gathered}$

Now, for the general solution, it is known that for

$\begin{gathered} \cot (\frac{\pi}{6})=\sqrt[]{3} \\ \text{Then,} \\ \theta=\frac{2\pi}{3}+\pi\cdot n \end{gathered}$