## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 5 - Section 5.5 - Trigonometric Equations - Exercise Set - Page 703: 34

#### Answer

The solution in the interval $[0,2\pi )$ is $\frac{3\pi }{2}$.

#### Work Step by Step

We know that the period of the cosine function is $2\pi$. So, in the interval $(0,\,\,\pi ]$, the only value for which the cosine function is $-1$ is $\pi$. Therefore, all the solutions to $\cos \frac{2\theta }{3}=-1$ are given by: \begin{align} & \frac{2\theta }{3}=\pi +2n\pi \\ & \theta =\frac{3\pi }{2}+3n\pi \end{align} Where, n is any integer. And the solution in the interval $[0,2\pi )$ is obtained by letting $n=0$. Thus, the equation is calculated by taking first $n$ as 0. It can be further simplified as follows. \begin{align} & \theta =\frac{3\pi }{2}+3n\pi \\ & =\frac{3\pi }{2}+3\times 0\times \pi \\ & =\frac{3\pi }{2}+0 \\ & =\frac{3\pi }{2} \end{align}

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