Answer
True.
Work Step by Step
Please note that vertical asymptotes of $y=2 \csc 2x=\frac{2}{\sin 2x}$ occur at $x$-intercepts of $y=2\sin 2x$. So $x=\frac{\pi }{2}$ is a vertical asymptote of $y=2 \csc 2x$ since $2\sin 2 (\frac{\pi }{2})=0$.
Precalculus (6th Edition) Blitzer
by Blitzer, Robert F.
Published by
Pearson
ISBN 10:
0-13446-914-3
ISBN 13:
978-0-13446-914-0
Chapter 4 - Section 4.6 - Graphs of Other Trigonometric Functions - Concept and Vocabulary Check - Page 609: 8
Update this answer!
You can help us out by revising, improving and updating this answer.
Update this answerAfter 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.
