## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 5 - Section 5.5 - Trigonometric Equations - Exercise Set - Page 704: 112

#### Answer

The solution is $\underline{x=1.3770,4.9062}.$

#### Work Step by Step

We know that for a standard quadratic equation of the form $a{{x}^{2}}+bx+c=0$, the solution is: $x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ And replace $x\text{ by }\cos x\text{ }$ in the standard quadratic equation to obtain the given equation in $\cos x$. Then the solution of the given equation is: \begin{align} & \cos x=\frac{-\left( 5 \right)\pm \sqrt{{{\left( 5 \right)}^{2}}-\left( 4\times 1\times -1 \right)}}{2\times 1} \\ & \cos x=\frac{-5\pm \sqrt{25+4}}{2} \\ \end{align} After that, simplify the equation: \begin{align} & \cos x=\frac{-5+\sqrt{29}}{2} \\ & \cos x\approx 0.1926 \\ \end{align} And the solution for addition will be: \begin{align} & \cos x=\frac{-5-\sqrt{29}}{2} \\ & \cos x\approx -5.1926 \\ \end{align} We know that $\operatorname{cosine}$ of any number lies only in the interval $\left[ -1,1 \right]$. So $\cos x\ne -5.1926$ and it is not the solution. The only solution is: \begin{align} & \cos x\approx 0.1926 \\ & x\approx {{\cos }^{-1}}\left( 0.1926 \right) \end{align} We have to calculate the equation using a calculator; it is required to keep the calculator in radian mode and find the value of ${{\cos }^{-1}}\left( 0.1926 \right)$ \begin{align} & x\approx {{\cos }^{-1}}\left( 0.1926 \right) \\ & \approx 1.3770 \end{align} And the cosine value is positive in I and IV quadrants. So $x\approx 1.3770$ Thus, the other solution will be: \begin{align} & x\approx 2\pi -1.3770 \\ & \approx 2\times 3.14159-1.3770 \\ & \approx 4.9062 \end{align}

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.