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 6.3 Trigonometric Equations II - 6.3 Exercises - Page 280: 56

Answer

$x = 0.692$ $x = 2.082$
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Work Step by Step

$3cos~\frac{x}{2}+\sqrt{x}-2 = -\frac{x}{2}+2$ We can graph the two equations $~~y = 3cos~\frac{x}{2}+\sqrt{x}-2 ~~$ and $~~y = -\frac{x}{2}+2~~$ to find the points of intersection. The blue graph is $~~y = -\frac{x}{2}+2~~$ The red graph is $~~y = 3cos~\frac{x}{2}+\sqrt{x}-2 ~~$ On the interval $[0,2\pi]$, we can see that the points of intersection occur when $~~x = 0.692~~$ and $~~x = 2.082$ The solution over the given interval is: $x = 0.692$ $x = 2.082$
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