Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 6 - Inverse Circular Functions and Trigonometric Equations - Section 6.2 Trigonometric Equations I - 6.2 Exercises - Page 274: 64

Answer

The two solutions in the given interval are: $x = 0$ $x = 0.3760$
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Work Step by Step

$x^3-cos^2~x = \frac{x}{2}-1$ $cos^2~x = x^3-\frac{x}{2}+1$ We can graph the two equations $~~y = cos^2~x~~$ and $~~y = x^3-\frac{x}{2}+1~~$ to find the points of intersection. The blue graph is $~~y = x^3-\frac{x}{2}+1$ The red graph is $~~y = cos^2~x$ We can see that the points of intersection occur when $~~x = 0~~$ and $~~x = 0.3760$ The two solutions in the given interval are: $x = 0$ $x = 0.3760$
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