Trigonometry 7th Edition

Published by Cengage Learning
ISBN 10: 1111826854
ISBN 13: 978-1-11182-685-7

Chapter 6 - Section 6.2 - More on Trigonometric Equations - 6.2 Problem Set - Page 332: 24



Work Step by Step

$4cos^2(x)-4sin(x)-5=0$ $4-4sin^2(x)-4sin(x)-5=0\;\;\;\;\;\;\;\;\;\;$ $4sin^2(x)+4sin(x)+1=0\;\;\;\;\;\;\;\;\;\;\;\;$ $cos(x)=\frac{-(4)\pm \sqrt{(4)^2-(4.4.(1))}}{2.4}=\frac{-1}{2}$ $sin(x)=\frac{-1}{2}$ $x=sin^{-1}(\frac{-1}{2})$ We know $ sin(x) $ is negative in quadrant $III$ and quadrant $IV$ $x=\pi +\frac{\pi}{6}\;\;\;\;\;\;\;\;or\;\;\;\;\;\;\;\;x=2\pi-\frac{\pi}{6}$ $x=\frac{7\pi}{6}\;\;\;\;\;\;\;\;\;\;\;\;\;\;or\;\;\;\;\;\;\;\;\;\;\;x=\frac{11\pi}{6}$ $x=\{\frac{7\pi}{6},\frac{11\pi}{6}\}$
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.