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: 45


$$x=\frac{\pi }{2}, \qquad x=\frac{3\pi }{2}$$

Work Step by Step

We solve by factoring: $$\sin ^2 \theta -1=0 \quad \Rightarrow \quad \sin ^2 \theta =1 \quad \Rightarrow \quad \sin \theta = \pm 1 \quad \Rightarrow \quad x=\frac{\pi }{2}+ 2n\pi, \quad \text{ or } \quad x=\frac{3\pi }{2}+ 2n \pi, \quad n \in \mathbb{Z}$$So, $x= \frac{\pi }{2}$ and $x= \frac{3\pi }{2}$ are the only solutions in the interval $[0, 2\pi )$.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.