Question

I need help with this practice problem solving It is possible to choose more than one answer option

Answer 1

We have to find the general solution to:

`3\cot \theta=-\sqrt[]{3}`

We can start rearranging the equation:

`\begin{gathered} \cot \theta=-\frac{\sqrt[]{3}}{3} \\ (1)/(\tan\theta)=-\frac{\sqrt[]{3}}{3} \\ (1)/(\tan\theta)=-\frac{1}{\sqrt[]{3}} \\ \tan \theta=-\sqrt[]{3} \end{gathered}`

We will have one solution per cycle of length π.

We calculate the solution for the first period as:

`\begin{gathered} \tan \theta=-\sqrt[]{3} \\ \theta=\arctan (-\sqrt[]{3}) \\ \theta=(2\pi)/(3) \end{gathered}`

We then can generalize for the other periods as:

`\theta=(2\pi)/(3)+n\pi`

Answer: θ = 2π/3 + πn