## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 5 - Section 5.5 - Trigonometric Equations - Exercise Set - Page 704: 75

#### Answer

$\{\frac{\pi}{8},\frac{3\pi}{8},\frac{9\pi}{8},\frac{11\pi}{8} \}$

#### Work Step by Step

Step 1. Using the identity $sin(2x)=2sin(x)cos(x)$, we have $sin(2x)=\frac{\sqrt 2}{2}$, which gives $2x=2k\pi+\frac{\pi}{4}$ and $2x=2k\pi+\frac{3\pi}{4}$ or $x=k\pi+\frac{\pi}{8}$ and $x=k\pi+\frac{3\pi}{8}$ where $k$ is an integer. Step 2. For $x=k\pi+\frac{\pi}{8}$, we can find all x-values in $[0,2\pi)$ as $x=\frac{\pi}{8},\frac{9\pi}{8}$ Step 3. For $x=k\pi+\frac{3\pi}{8}$, we can find all x-values in $[0,2\pi)$ as $x=\frac{3\pi}{8},\frac{11\pi}{8}$ Step 4. The solutions for the original equation in $[0,2\pi)$ are $\{\frac{\pi}{8},\frac{3\pi}{8},\frac{9\pi}{8},\frac{11\pi}{8} \}$

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