Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 1: Functions - Section 1.3 - Trigonometric Functions - Exercises 1.3 - Page 28: 52

Answer

$\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}$

Work Step by Step

From $ sin^2\theta=cos^2\theta $, we have $ sin^2\theta=1-sin^2\theta $ and thus $ sin^2\theta=\frac{1}{2}$. This leads to $ sin\theta=\frac{\sqrt 2}{2}$ and $ sin\theta=-\frac{\sqrt 2}{2}$. Within $[0,2\pi]$, the solutions are: $\theta=\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}$
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