Precalculus (6th Edition) Blitzer

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

Chapter 5 - Section 5.5 - Trigonometric Equations - Concept and Vocabulary Check - Page 703: 6


If $2{{\cos }^{2}}x-9\cos x-5=0$, then $\underline{2\cos x+1}=0$ or $\underline{\cos x-5}=0$. Of these two equations, the equation that has no solution is $\cos x-5=0$.

Work Step by Step

To evaluate another value, the middle term splitting is used: $\begin{align} & 2{{\cos }^{2}}x-9\cos x-5=0 \\ & 2{{\cos }^{2}}x+\cos x-10\cos x-5=0 \\ & \cos x\left( 2\cos x+1 \right)-5\left( 2\cos x+1 \right)=0 \\ & \left( 2\cos x+1 \right)\left( \cos x-5 \right)=0 \end{align}$ Therefore, the values will be $2\cos x+1$, another value will be $\cos x-5$ and the equation that has no solution will be $\cos x-5=0$. Thus, the values will be $2\cos x+1,\,\cos x-5\,\text{ and }\,\cos x-5=0$.
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