Precalculus (6th Edition) Blitzer

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

Chapter 5 - Section 5.5 - Trigonometric Equations - Exercise Set - Page 705: 153


The statement makes sense.

Work Step by Step

Similarities: Solve $4x+1=3$ by isolating $x$ on one side of the equation: $\begin{align} & 4x+1=3 \\ & 4x=2 \\ & x=\frac{1}{2} \end{align}$ Isolate the $\sin \theta $ from the equation $4\sin \theta +1=3$ $\begin{align} & 4\sin \theta =2 \\ & \sin \theta =\frac{1}{2} \\ \end{align}$ Differences: Isolate the $x$ and solve to get the absolute value for $x$. However, when $\sin \theta $ is isolated, solve for the value of $\theta $ for which $\sin \theta =\frac{1}{2}$; that is, $\sin \theta =\sin \left( \frac{\pi }{6} \right)$ or $\sin \theta =\sin \left( \frac{5\pi }{6} \right)$. So, the values of $\theta $ are $\frac{\pi }{6}$ and $\frac{5\pi }{6}$ between $[0,2\pi )$. Therefore, the statement makes sense.
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