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: 39

Answer

$(1,\frac{\pi}{12}), (1,\frac{5\pi}{12}),(1,\frac{13\pi}{12}),(1,\frac{17\pi}{12}),(-1,\frac{7\pi}{12}),(-1,\frac{11\pi}{12}),(-1,\frac{19\pi}{12}),(-1,\frac{23\pi}{12})$

Work Step by Step

$2sin2\theta=1$ $sin2\theta=\frac{1}{2}$ $\theta=\frac{\pi}{12},\frac{5\pi}{12}$ Now, $2sin2\theta=-1$ $sin2\theta=\frac{1}{2}$ $\theta=\frac{7\pi}{12},\frac{11\pi}{12}$ Therefore, the point of intersections are: $(1,\frac{\pi}{12}), (1,\frac{5\pi}{12}),(1,\frac{13\pi}{12}),(1,\frac{17\pi}{12}),(-1,\frac{7\pi}{12}),(-1,\frac{11\pi}{12}),(-1,\frac{19\pi}{12}),(-1,\frac{23\pi}{12})$
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