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: 53



Work Step by Step

Since $ sin2\theta=2sin\theta\cdot cos\theta $, we can rewrite the equation as $2sin\theta\cdot cos\theta-cos\theta=0$, or $ cos\theta(2sin\theta-1)=0$. The solutions are $ cos\theta=0$ or $2sin\theta-1=0$ which is $ sin\theta=\frac{1}{2}$ Within $[0,2\pi]$, we have $\theta=\frac{\pi}{2},\frac{3\pi}{2},\frac{\pi}{6},\frac{5\pi}{6}$
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