Precalculus (6th Edition) Blitzer

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

Chapter 6 - Cumulative Review Exercises - Page 801: 4


The solutions are $\theta =\frac{3\pi }{4},\frac{7\pi }{4}$.

Work Step by Step

Solve equation (1): $\begin{align} & \sin \theta \cos \theta =-\frac{1}{2} \\ & \frac{\sin 2\theta }{2}=-\frac{1}{2} \\ & \sin 2\theta =-1 \end{align}$ As the given period is $2\pi $, therefore the value on which $\sin \theta =-1$ is $\frac{3\pi }{2}$. Now, $2\theta =\frac{3\pi }{2}$. As the given period is $2\pi $, all the solutions to $\sin 2\theta =-1$ are: $\begin{align} & 2\theta =\frac{3\pi }{2}+2n\pi \\ & \theta =\frac{3\pi }{4}+n\pi \end{align}$ Therefore, the solution in $\left[ 0,\left. 2\pi \right) \right.$ can be obtained by letting $ n=0$ and $ n=1$, such that: $\theta =\frac{3\pi }{4},\frac{7\pi }{4}$
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