Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 5 - Section 5.5 - Trigonometric Equations - Exercise Set - Page 703: 35

Answer

The solutions in the interval $[0,2\pi )$ are $\frac{4\pi }{9}$, $\frac{8\pi }{9}$ and $\frac{16\pi }{9}$.

Work Step by Step

We know that the period of the secant function is $2\pi $. In the interval $\left[ 0,\left. 2\pi \right) \right.$, there are two values at which the secant function is $-2$. One is $\frac{2\pi }{3}$ and the secant is negative in quadrant II, thus the other value is: $\begin{align} & 2\pi -\frac{2\pi }{3}=\frac{6\pi -2\pi }{3} \\ & =\frac{4\pi }{3} \end{align}$ Therefore, all the solutions to $\sec \frac{3\theta }{2}=-2$ are given by: $\begin{align} & \frac{3\theta }{2}=\frac{2\pi }{3}+2n\pi \\ & \theta =\frac{4\pi }{9}+\frac{4n\pi }{3} \end{align}$ Or, $\begin{align} & \frac{3\theta }{2}=\frac{4\pi }{3}+2n\pi \\ & \theta =\frac{8\pi }{9}+\frac{4n\pi }{3} \end{align}$ Where, n is any integer. And the solutions in the interval $[0,2\pi )$ are obtained by letting $n=0$ and $n=1$. Thus, the equation is calculated by taking first $n$ as 0 and then as 1. It can be further simplified as follows. $\begin{align} & \theta =\frac{4\pi }{9}+\frac{4n\pi }{3} \\ & =\frac{4\pi }{9}+\frac{4\times 0\times \pi }{3} \\ & =\frac{4\pi }{9}+0 \\ & =\frac{4\pi }{9} \end{align}$ $\begin{align} & \theta =\frac{4\pi }{9}+\frac{4n\pi }{3} \\ & =\frac{4\pi }{9}+\frac{4\times 1\times \pi }{3} \\ & =\frac{4\pi }{9}+\frac{4\pi }{3} \\ & =\frac{4\pi +12\pi }{9} \end{align}$ $=\frac{16\pi }{9}$ The second equation is also computed by taking first $n$ as 0 and then as 1. It can be further simplified as follows. $\begin{align} & \theta =\frac{8\pi }{9}+\frac{4n\pi }{3} \\ & =\frac{8\pi }{9}+\frac{4\times 0\times \pi }{3} \\ & =\frac{8\pi }{9}+0 \\ & =\frac{8\pi }{9} \end{align}$ $\begin{align} & \theta =\frac{8\pi }{9}+\frac{4n\pi }{3} \\ & =\frac{8\pi }{9}+\frac{4\times 1\times \pi }{3} \\ & =\frac{8\pi }{9}+\frac{4\pi }{3} \\ & =\frac{8\pi +12\pi }{9} \end{align}$ $=\frac{20\pi }{9}$ Since $\frac{20\pi }{9}$ is not in $[0,2\pi )$, the actual solutions in the interval $[0,2\pi )$ will be $\frac{4\pi }{9}$, $\frac{8\pi }{9}$, and $\frac{16\pi }{9}$.
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