Question

find all solutions to the equation.

\\( \cos \theta = - \frac { \sqrt { 3 } } { 2 } \\)

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: Recall the reference angle

We know that \(\cos\theta=\frac{\sqrt{3}}{2}\) when \(\theta = \frac{\pi}{6}\).

Step2: Determine the quadrants for negative cosine

Since \(\cos\theta=-\frac{\sqrt{3}}{2}<0\), \(\theta\) is in the second or third quadrant.
In the second quadrant, \(\theta=\pi-\frac{\pi}{6}=\frac{5\pi}{6}\).
In the third quadrant, \(\theta=\pi+\frac{\pi}{6}=\frac{7\pi}{6}\).

Step3: Add the period of the cosine function

The period of \(y = \cos\theta\) is \(2\pi\). So the general solutions are \(\theta=\frac{5\pi}{6}+2k\pi,k\in\mathbb{Z}\) or \(\theta=\frac{7\pi}{6}+2k\pi,k\in\mathbb{Z}\)

Answer:

\(\theta=\frac{5\pi}{6}+2k\pi,k\in\mathbb{Z}\text{ or }\theta=\frac{7\pi}{6}+2k\pi,k\in\mathbb{Z}\)