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: 22

Answer

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

Work Step by Step

$y=-\sqrt 3+2csc\frac{x}{2}$ $y+\sqrt 3=2csc\frac{x}{2}$ 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})]$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.