Question

find all solutions to the equation.

\\( \cot \theta + \sqrt { 3 } = 0 \\)

write your answer in radians in terms of \\( \pi \\), and use the \or\ button as necessary

example: \\( \theta = \frac { \pi } { 5 } + 2 k \pi, k \in \mathbb { z } \\) or \\( \theta = \frac { \pi } { 7 } + k \pi, k \in \mathbb { z } \\)

\\( \theta = \square \\)

Explanation:

Step1: Isolate the cotangent function

Given the equation \(\cot\theta+\sqrt{3} = 0\), we can rewrite it as \(\cot\theta=-\sqrt{3}\).
Since \(\cot\theta=\frac{\cos\theta}{\sin\theta}\), and we know that \(\cot\theta =-\sqrt{3}\), also, we know that \(\cot\theta=\frac{1}{\tan\theta}\), so \(\tan\theta=-\frac{1}{\sqrt{3}}\).

Step2: Find the reference angle

We know that \(\tan\frac{\pi}{6}=\frac{1}{\sqrt{3}}\). The tangent function \(y = \tan\theta\) has a period of \(\pi\).
The tangent function is negative in the second and fourth quadrants.
For \(\tan\theta=-\frac{1}{\sqrt{3}}\), in the second - quadrant \(\theta=\pi-\frac{\pi}{6}=\frac{5\pi}{6}\), and in the fourth - quadrant \(\theta = 2\pi-\frac{\pi}{6}=\frac{11\pi}{6}\).
The general solution of the tangent function \(y=\tan\theta\) is given by \(\theta=\theta_{0}+k\pi\), where \(\theta_{0}\) is a particular solution and \(k\in\mathbb{Z}\) (the set of integers).
Since the period of \(y = \cot\theta\) (or \(y=\tan\theta\)) is \(\pi\), the general solution of \(\cot\theta=-\sqrt{3}\) (or \(\tan\theta =-\frac{1}{\sqrt{3}}\)) is \(\theta=\frac{5\pi}{6}+k\pi,k\in\mathbb{Z}\)

Answer:

\(\theta=\frac{5\pi}{6}+k\pi,k\in\mathbb{Z}\)