Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 6 - Inverse Circular Functions and Trigonometric Equations - Section 6.2 Trigonometric Equations I - 6.2 Exercises - Page 268: 59

Answer

The two solutions over the given interval are: $x = 0.6806$ $x = 1.4159$

Work Step by Step

$x^2+sin~x-x^3-cos~x = 0$ $sin~x-cos~x = x^3-x^2$ We can graph the two equations $~~y = sin~x-cos~x~~$ and $~~y = x^3-x^2~~$ to find the points of intersection. The blue graph is $~~y = x^3-x^2~~$ The red graph is $~~y = sin~x-cos~x~~$ We can see that the points of intersection occur when $~~x = 0.6806~~$ and $~~x = 1.4159$ The two solutions over the given interval are: $x = 0.6806$ $x = 1.4159$
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