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 6.4 Equations Involving Inverse Trigonometric Functions - 6.4 Exercises - Page 286: 23


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

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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