Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 8 - Application of Trigonometry - 8.7 Polar Equations and Graphs - 8.7 Exercises - Page 819: 4

Answer

$\pm 1$

Work Step by Step

$r^{2}=-2\cos 2\theta \Rightarrow r^{2}=-2\cos \left( 2\times 60\right) =-2\cos 120=-2\times \left( -\dfrac {1}{2}\right) =1\Rightarrow r=\pm 1$
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