Chapter 6 - Inverse Circular Functions and Trigonometric Equations - Section 6.4 Equations Involving Inverse Trigonometric Functions - 6.4 Exercises - Page 280: 30

$\frac{3\sqrt 3+2\pi}{6}$

Given: $arccos({y-\frac{\pi}{3}})=\frac{\pi}{6}$ $({y-\frac{\pi}{3}})=cos(\frac{\pi}{6})$ Apply definition of cosine. $({y-\frac{\pi}{3}})=\frac{\sqrt 3}{2}$ $y=\frac{\sqrt 3}{2}+\frac{\pi}{3}$ $y=\frac{3\sqrt 3+2\pi}{6}$ Hence, $y=\frac{3\sqrt 3+2\pi}{6}$