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 6.4 Equations Involving Inverse Trigonometric Functions - 6.4 Exercises - Page 280: 21

Answer

$x=\frac{1}{2}sec^{-1}(\frac{y-\sqrt 2}{3})$

Work Step by Step

$y=\sqrt 2+3sec2x$ $y-\sqrt 2= 3sec2x$ Divide by 3 on both sides. $ sec2x=\frac{y-\sqrt 2}{3}$ $x=\frac{1}{2}sec^{-1}(\frac{y-\sqrt 2}{3})$ Hence, $x=\frac{1}{2}sec^{-1}(\frac{y-\sqrt 2}{3})$
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