Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 10 - Section 10.2 - Areas and Lengths in Polar Coordinates - 10.4 Exercises - Page 673: 40

Answer

The intersection points are: $(\frac{1}{\sqrt{2}},\pi/12), (\frac{-1}{\sqrt{2}},5\pi/12), (\frac{1}{\sqrt{2}},3\pi/4)$

Work Step by Step

$sin3\theta=cos3\theta$ $\frac{sin3\theta}{cos3\theta}=1$ $tan3\theta=1$ The periodicity of $tan3\theta$ is $\frac{\pi}{3}$ Therefore, the points of intersection are: $\theta=\frac{\pi}{12}+n\frac{\pi}{3}$ $\theta$ can take on the values: $\frac{\pi}{12}, \frac{5\pi}{12},\frac{9\pi}{12}$ The intersection points are: $(\frac{1}{\sqrt{2}},\pi/12), (\frac{-1}{\sqrt{2}},5\pi/12), (\frac{1}{\sqrt{2}},3\pi/4)$
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