Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 12 - Sequences; Induction; the Binomial Theorem - Chapter Review - Cumulative Review - Page 840: 10



Work Step by Step

We have to solve the equation: $2\sin^2x-\sin x-3=0$, $0\leq x<2\pi$ Factor (or note $y=\sin x$ and use the quadratic formula): $2\sin^2 x+2\sin x-3\sin x-3=0$ $2\sin x(\sin x+1)-3(\sin x+1)=0$ $(\sin x+1)(2\sin x-3)=0$ Solve for $x$: $\sin x+1=0$ or $2\sin x-3=0$ $\sin x=-1$ or $\sin x=\dfrac{3}{2}$ $x=\dfrac{3\pi}{2}$ (the equation $\sin x=\dfrac{3}{2}$ has no solution as $\sin x\leq 1$ The solution set is: $\left\{\dfrac{3\pi}{2}\right\}$
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