Precalculus (6th Edition) Blitzer

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

Chapter 8 - Cumulative Review Exercises - Page 952: 24


The solution is $x=\frac{3\pi }{2}$.

Work Step by Step

Consider the given trigonometric equation: $\begin{align} & {{\cos }^{2}}x+\sin x+1=0 \\ & 1-{{\sin }^{2}}x+\sin x+1=0 \\ & {{\sin }^{2}}x-\sin x-2=0 \\ & {{\sin }^{2}}x+\sin x-2\sin x-2=0 \end{align}$ It can be further simplified as: $\begin{align} & \sin x\left( \sin x+1 \right)-2\left( \sin x+1 \right)=0 \\ & \left( \sin x+1 \right)\left( \sin x-2 \right)=0 \end{align}$ Therefore, $\sin x=2\,\,\text{or}\,\,-1$ As $-1\le \sin x\le 1$ hence $\sin x\ne 2$ Therefore, $\sin x=-1$ That is $\begin{align} & \sin x=\sin \frac{\pi }{2} \\ & x=\frac{3\pi }{2} \end{align}$ The solution of the equation in $\left[ 0,2\pi \right]$ is $\frac{3\pi }{2}$.
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