Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 1 - Functions - 1.4 Trigonometric Functions and Their Inverse - 1.4 Exercises - Page 48: 44

Answer

$$\theta = \frac{\pi }{2} + 2n\pi {\text{ and }}\theta = \frac{{3\pi }}{2} + 2n\pi ,\,\,\,\,n = 0, \pm 1, \pm 2, \ldots $$

Work Step by Step

$$\eqalign{ & {\sin ^2}\theta - 1 = 0 \cr & {\text{Add 1 to both sides of the equation}} \cr & {\sin ^2}\theta = 1 \cr & \sin \theta = \pm 1 \cr & \sin \theta = 1{\text{ or sin}}\theta = - 1 \cr & {\text{In the interval }}\left[ {0,2\pi } \right)\,\,\sin \theta = 1{\text{ for }}\theta = \frac{\pi }{2}{\text{ and sin}}\theta = - 1{\text{ for}} \cr & \theta = \frac{{3\pi }}{2} \cr & {\text{Moreover, because }}\sin \theta {\text{ has a period of 2}}\pi ,{\text{ there are infinitely }} \cr & {\text{many other solutions, which can be written as}} \cr & \theta = \frac{\pi }{2} + 2n\pi {\text{ and }}\theta = \frac{{3\pi }}{2} + 2n\pi \cr & {\text{Where }}n{\text{ is an integer}}{\text{.}} \cr & {\text{The general solution is:}} \cr & \theta = \frac{\pi }{2} + 2n\pi {\text{ and }}\theta = \frac{{3\pi }}{2} + 2n\pi ,\,\,\,\,n = 0, \pm 1, \pm 2, \ldots \cr} $$
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