Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 7 - Trigonometric Identities and Equations - 7.6 Trigonometric Equations - 7.6 Exercises - Page 720: 26

Answer

$0$, $\frac{2\pi}{3}$, $\frac{4\pi}{3}$

Work Step by Step

$2\cos^2 x-\cos x=1$ $2\cos^2 x-\cos x-1=0$ $(2\cos x+1)(\cos x-1)=0$ $2\cos x+1=0$ or $\cos x-1=0$ If $2\cos x+1=0$: $2\cos x=-1$ $\cos x=-\frac{1}{2}$ The only solutions in $[0, 2\pi)$ are $\frac{2\pi}{3}$ and $\frac{4\pi}{3}$. If $\cos x-1=0$: $\cos x=1$ The only solution in $[0, 2\pi)$ is $0$.
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.