Chapter 6 - Inverse Circular Functions and Trigonometric Equations - Section 6.4 Equations Involving Inverse Trigonometric Functions - 6.4 Exercises - Page 280: 22

$ x=2csc^{-1}[(\frac{y+\sqrt 3}{2})]$

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