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 704: 107


The solutions to the given equation are $\frac{\pi }{6},\frac{\pi }{2},\frac{5\pi }{6},\text{ and }\ \frac{3\pi }{2}$.

Work Step by Step

We have to solve the equation on the interval $[0,2\pi )$: $\begin{align} & \text{cos}\ x\ \text{csc}\ x=2\text{cos}\ x \\ & \text{cos}\ x\ \text{csc}\ x-2\text{cos}\ x=0 \\ & cos\ x\left( \text{csc}\ x-2 \right)=0 \end{align}$ And each factor needs to be calculated as: $\text{cos}\ x=0\ $ or $\begin{align} & \text{csc}\ x-2=0 \\ & \csc \ x=0+2 \\ & =2 \end{align}$ Then, solve for $x$ on the interval $[0,2\pi )$. So, in the quadrant graph, the value of cosine is $0$ at $\frac{\pi }{2}$ and $\frac{3\pi }{2}$. This implies, $\begin{align} & \text{cos}x=\cos \frac{\pi }{2} \\ & x=\ \frac{\pi }{2} \end{align}$ $\begin{align} & \text{cos}x=\cos \frac{3\pi }{2} \\ & x=\ \frac{3\pi }{2} \end{align}$ And, The value of cosecant is not shown in the quadrant graph. By using the reciprocal identity of trigonometry, $\csc x=\frac{1}{\sin x}$, the cosecant function gets converted into the sine function as, $\begin{align} & \frac{1}{\sin x}=2 \\ & \sin x=\frac{1}{2} \end{align}$ So, in the quadrant graph, the value of sine is $\frac{1}{2}$ at $\frac{\pi }{6}$ and $\frac{5\pi }{6}$. This implies, $\begin{align} & \sin x=\sin \frac{\pi }{6} \\ & x=\frac{\pi }{6} \end{align}$ $\begin{align} & \sin x=\sin \frac{5\pi }{6} \\ & x=\frac{5\pi }{6} \end{align}$ These are the proposed solutions of the cosine and sine functions. Thus, the actual solutions in the interval $[0,2\pi )$ will be $\frac{\pi }{6},\frac{\pi }{2},\frac{5\pi }{6},\text{ and }\ \frac{3\pi }{2}$.
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