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 - 14.5 Write Trigonometric Functions and Models - 14.5 Exercises - Quiz for Lessons 14.3-14.5 - Page 945: 9


$x =n \pi +(-1)^{n+1} \dfrac{\pi}{6}$ and $x =n \pi + (-1)^n \dfrac{\pi}{2}$

Work Step by Step

Re-arrange the given equation as: $ 2 \sin^2 x-\sin x=1$ or, $(2\sin x+1)(\sin x-1)=0$ or, $ 2 (\sin x+\dfrac{1}{2})(\sin x-1)=0$ When $ \sin x=\dfrac{-1}{2}$ Therefore, the general solution of $\sin x$ is: $x =n \pi +(-1)^{n+1} \dfrac{\pi}{6}$ when $\sin x=1$ Therefore, the general solution of $\sin x$ is: $x =n \pi + (-1)^n \dfrac{\pi}{2}$
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.