Algebra 2 (1st Edition)

Published by McDougal Littell
ISBN 10: 0618595414
ISBN 13: 978-0-61859-541-9

Chapter 14 Trigonometric Graphs, Identities, and Equations - 14.6 Apply Sum and Difference Formulas - 14.6 Exercises - Skill Practice - Page 952: 32


Option C

Work Step by Step

Here, we have $\sin (x- 2\pi) +tan (x-2 \pi)=0$ The periodicity for $\sin x$ is $ 2 \pi$ and the periodicity for $\tan x$ is $ \pi$. $\sin x +\tan x=0 \implies \sin x+\dfrac{\sin x}{\cos x}=0$ or, $\sin x(1+\dfrac{1}{\cos x})=0$ Thus, $\sin x=0$ or, $(1+\dfrac{1}{\cos x})=0$ We can see that the interval for $\sin x$ is $0$ when $x= 2\pi$ Hence, Option C is correct.
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.