Precalculus (6th Edition) Blitzer

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

Chapter 7 - Cumulative Review Exercises - Page 880: 38


The solution of the given trigonometric equation is $\theta =0,\pi$

Work Step by Step

We have the equation as $\sin \theta =\tan \theta $. By solving the equation, $\begin{align} & \sin \theta =\tan \theta \\ & \sin \theta =\frac{\sin \theta }{\cos \theta } \\ & \sin \theta =\frac{\sin \theta }{\sqrt{1-{{\sin }^{2}}\theta }} \\ & \sin \theta \times \sqrt{1-{{\sin }^{2}}\theta }=\sin \theta \end{align}$ Then, $\begin{align} & \sin \theta \times \sqrt{1-{{\sin }^{2}}\theta }-\sin \theta =0 \\ & \sin \theta \left( \sqrt{1-{{\sin }^{2}}\theta }-1 \right)=0 \end{align}$ And from the two factors, we get $\begin{align} & \sin \theta =0 \\ & \theta =n\pi \end{align}$ and, $\begin{align} & 1-{{\sin }^{2}}\theta =1 \\ & {{\sin }^{2}}\theta =0 \\ & \sin \theta =0 \\ & \theta =n\pi \end{align}$ Therefore, we get $\theta =n\pi $. The provided range is $\left[ 0,2\pi \right)$. Hence, $\theta =0,\pi $ are the solutions of the provided equation.
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