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 - 14.7 Apply Double-Angle and Half-Angle Formulas - 14.7 Exercises - Skill Practice - Page 960: 44


$x=\pi +2 \pi n; x= \dfrac{\pi}{3}+2 \pi n ; x= \dfrac{ 5 \pi}{3}+2 \pi n$ and $x=2 \pi n$

Work Step by Step

Here, we have $ 2 \sin x \cos x=\sin x$ It has to be noted that $\sin x (2 \cos x-1)=0$ This gives: $\sin x=0 \implies x=\pi, 2 \pi$ The general solution for $\cos (x/2)=\cos (0)$ is $ (x/2)=2 n \pi \pm (0)$ $ 2 \cos x-1=0 \implies \cos x=\dfrac{1}{2}$ This gives: $ x=\dfrac{\pi}{3}; \dfrac{ 5\pi}{3}$ Hence, we have $x=\pi +2 \pi n; x= \dfrac{\pi}{3}+2 \pi n ; x= \dfrac{ 5 \pi}{3}+2 \pi n$ and $x=2 \pi n$
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.