## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 5 - Section 5.5 - Trigonometric Equations - Exercise Set - Page 704: 67

#### Answer

$\{\frac{\pi}{6},\frac{5\pi}{6} \}$

#### Work Step by Step

Step 1. Use the identity $cos^2x=1-sin^2x$, we have $4-4sin^2x+4sin(x)-5=0$ or $4sin^2x-4sin(x)+1=0$, or $(2sin(x)-1)^2=0$, which gives solutions as $sin(x)=\frac{1}{2}$ (multiplicity 2) Step 2. For $sin(x)=\frac{1}{2}$, we can find all x-values in $[0,2\pi)$ as $x=\frac{\pi}{6},\frac{5\pi}{6}$ Step 3. The solutions for the original equation in $[0,2\pi)$ are $\{\frac{\pi}{6},\frac{5\pi}{6} \}$

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.