Precalculus (6th Edition) Blitzer

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

Chapter 10 - Section 10.6 - Counting Principles, Permutations, and Combinations - Exercise Set - Page 1105: 98


The solutions are $\frac{\pi }{2},\frac{\pi }{6},\frac{5}{6}$.

Work Step by Step

Let us consider the provided equation, $\cos 2x+3\sin x-2=0$ Since, $\cos 2x=1-2{{\sin }^{2}}x $ So, $\begin{align} & \cos 2x+3\sin x-2=0 \\ & 1-2{{\sin }^{2}}x+3\sin x-2=0 \end{align}$ Multiply both sides by $-1$, to get $\begin{align} & 2{{\sin }^{2}}x+3\sin x+1=0 \\ & \left( 2\sin x-1 \right)\left( \sin x-1 \right)=0 \end{align}$ It implies. $2\sin x-1=0$ Or $ sinx-1=0$ Now, $\begin{align} & 2\sin x-1=0 \\ & \sin x=\frac{1}{2} \\ & x=\frac{\pi }{6},\frac{5\pi }{6} \end{align}$ And, $\begin{align} & sinx-1=0 \\ & \sin x=1 \\ & x=\frac{\pi }{2} \end{align}$ Thus, the solutions are $\frac{\pi }{2},\frac{\pi }{6},\frac{5}{6}$.
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