Precalculus (10th Edition)

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

Chapter 6 - Trigonometric Functions - Chapter Review - Cumulative Review - Page 437: 1

Answer

$\left\{-1,\dfrac{1}{2}\right\}$

Work Step by Step

First we factor: $(x+1)(2x-1)=0$ Then use the Zero Factor property by equating each factor to zero: $x+1=0$ or $2x-1=0$ Solve the equations: $x=-1$ or $x=\dfrac{1}{2}$ The solution set is: $\left\{-1,\dfrac{1}{2}\right\}$
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