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: 48


For the provided equation, the value of x is $ x=\frac{\pi }{6},\frac{5\pi }{6},\frac{7\pi }{6},\frac{11\pi }{6}$.

Work Step by Step

We have the provided equation as $4{{\cos }^{2}}x=3$ Then, solving, $\begin{align} & 4{{\cos }^{2}}x=3 \\ & {{\cos }^{2}}x=\frac{3}{4} \\ & \cos x=\pm \frac{\sqrt{3}}{2} \end{align}$ For $\cos x=\frac{\sqrt{3}}{2}$ in the given range, $ x=\frac{\pi }{6},\frac{11\pi }{6}$ And for $\cos x=\frac{-\sqrt{3}}{2}$ $ x=\frac{5\pi }{6},\frac{7\pi }{6}$ Hence, we have $ x=\frac{\pi }{6},\frac{5\pi }{6},\frac{7\pi }{6},\frac{11\pi }{6}$.
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