Precalculus (6th Edition) Blitzer

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

Chapter 10 - Cumulative Review Exercises - Page 1128: 49


For the provided equation, the value of x is $ x=0,\frac{\pi }{3},\frac{5\pi }{3}$.

Work Step by Step

We have $2{{\sin }^{2}}x+3\cos x-3=0$ By solving, $\begin{align} & 2{{\sin }^{2}}x+3\cos x-3=0 \\ & 2\left( 1-{{\cos }^{2}}x \right)+3\cos x-3=0 \\ & 2-2{{\cos }^{2}}x+3\cos x-3=0 \\ & 2{{\cos }^{2}}x-3\cos x+1=0 \end{align}$ And factorizing: $\begin{align} & 2{{\cos }^{2}}x-2\cos x-\cos x+1=0 \\ & 2\cos x\left( \cos x-1 \right)-1\left( \cos x-1 \right)=0 \\ & \left( 2\cos x-1 \right)\left( \cos x-1 \right)=0 \\ & \cos x=1,\frac{1}{2} \end{align}$ And for $\begin{align} & \cos x=1 \\ & x=0 \end{align}$ And for $\begin{align} & \cos x=\frac{1}{2} \\ & x=\frac{\pi }{3},\frac{5\pi }{3} \end{align}$ Hence, $ x=0,\frac{\pi }{3},\frac{5\pi }{3}$.
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