Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 7 - Section 7.4 - Basic Trigonometric Equations - 7.4 Exercises - Page 569: 28

Answer

$\frac{\pi}{4}+2k\pi$, $\frac{7\pi}{4}+2k\pi$

Work Step by Step

$\begin{align} \sqrt{2}\cos\theta-1&=0\\ \sqrt{2}\cos\theta&=1\\ \cos\theta&=\frac{1}{\sqrt 2}\\ \cos\theta&=\frac{\sqrt{2}}{2} \end{align}$ Since $\cos \theta$ is positive in the first and fourth quadrants, the only values of $\theta$ between $0$ and $2\pi$ that satisfy the equation are $\frac{\pi}{4}$ and $\frac{7\pi}{4}$. Since $\cos \theta$ is a periodic function with period $2\pi$, we can add or subtract any multiple of $2\pi$ from $\frac{\pi}{4}$ or $\frac{7\pi}{4}$ and the equation would still be satisfied. So the answers are $\frac{\pi}{4}+2k\pi$ and $\frac{7\pi}{4}+2k\pi$ for any integer $k$.
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