Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 5 - Review Exercises - Page 708: 51


$2k\pi+\frac{\pi}{4}$, $2k\pi+\frac{3\pi}{4}$

Work Step by Step

Step 1. Given the equation $sin(x)=\frac{\sqrt 2}{2}$, we have to find the solutions in $[0,2\pi)$ as $x_1=\frac{\pi}{4}$ and $x_2=\frac{3\pi}{4}$ Step 2. Consider that the function has a period of $2\pi$; we can express the solutions as $x=2k\pi+\frac{\pi}{4}$ and $x=2k\pi+\frac{3\pi}{4}$ where $k$ is an integer.
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