## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 11 - Cumulative Review Exercises - Page 1180: 4

#### Answer

The solution of the equation ${{\cos }^{2}}x+\sin x+1=0$ is $\frac{3\pi }{2}$.

#### Work Step by Step

Consider the provided equation, ${{\cos }^{2}}x+\sin x+1=0$ Write ${{\cos }^{2}}x=1-{{\sin }^{2}}x$ using the identity ${{\cos }^{2}}x+{{\sin }^{2}}x=1$. \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=0 \end{align} Simplify it further as, \begin{align} & {{\sin }^{2}}x-\sin x-2=0 \\ & {{\sin }^{2}}x-2\sin x+\sin x-2=0 \\ & \left( \sin x-2 \right)\left( \sin x+1 \right)=0 \end{align} Equate the factors to zero to obtain the roots. $\sin x-2\ne 0$. The maximum value of $\sin x$ is 1. \begin{align} & \left( \sin x-2 \right)\left( \sin x+1 \right)=0 \\ & \sin x+1=0 \\ & \sin x=-1 \\ & x=\frac{3\pi }{2} \end{align} Thus, the solution of the equation ${{\cos }^{2}}x+\sin x+1=0$ is $x=\frac{3\pi }{2}$.

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