Precalculus (6th Edition) Blitzer

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

Chapter 6 - Cumulative Review Exercises - Page 801: 2


The solution of the equation $\sin \theta =\left( \frac{\pi }{6},\frac{5\pi }{6},\frac{\pi }{2} \right)$.

Work Step by Step

When the middle term will be split, the given equation will be: $\begin{align} & 2{{\sin }^{2}}\theta -3\sin \theta +1=0 \\ & \left( 2\sin \theta -1 \right)\left( \sin \theta -1 \right)=0 \end{align}$ Therefore, $\sin \theta =\frac{1}{2},1$ The values of $\sin \theta =\frac{1}{2}$ in the interval $\left[ 0,\left. 2\pi \right) \right.$ are $\frac{\pi }{6},\frac{5\pi }{6},\frac{\pi }{2}$.
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