Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 7 - Trigonometric Identities and Equations - 7.6 Trigonometric Equations - 7.6 Exercises - Page 720: 29



Work Step by Step

Substituting t$=\sin\theta$ ,the equation becomes $2t-1=\displaystyle \frac{1}{t}\qquad/\times t$ $2t^{2}-t=1$ $2t^{2}-t-1=0$ Using the quadratic formula, $t=\displaystyle \frac{1\pm\sqrt{1+8}}{4}=\frac{1\pm 3}{4}$ $t=1$ or $ t=-\displaystyle \frac{1}{2}$ Back substitute, use the unit circle: $\begin{array}{lll} \sin\theta=1 & or & \sin\theta=-\frac{1}{2}\\ \theta=90^{o} & or & \theta=210^{o},330^{0} \end{array}$ Solution set= $\{90^{o},210^{o},330^{o}\}$
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