Chapter 6 - Inverse Circular Functions and Trigonometric Equations - Section 6.4 Equations Involving Inverse Trigonometric Functions - 6.4 Exercises - Page 286: 23

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

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