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 - Exercise Set - Page 704: 96

Answer

The solution on the interval $[0,2\pi )$, correct to four decimal places is $0.4636$ , $2.6780$, $3.6052$ or $5.8195$.

Work Step by Step

Let us consider the provided expression: $5{{\sin }^{2}}x-1=0$ Add 1 on both sides: $\begin{align} & 5{{\sin }^{2}}x-1+1=0+1 \\ & 5{{\sin }^{2}}x=1 \end{align}$ And divide by 5 on both sides: $\begin{align} & \frac{5{{\sin }^{2}}x}{5}=\frac{1}{5} \\ & {{\sin }^{2}}x=\frac{1}{5} \end{align}$ And put the square root on both sides: $\begin{align} & \sin x=\pm \sqrt{\frac{1}{5}} \\ & \approx \pm 0.4472 \end{align}$ Since, the value of $\sin x\approx +0.4472$ $x\approx {{\sin }^{-1}}\left( +0.4472 \right)$ And the value of $x$ from $\sin x\approx -0.4472$ is: $x\approx {{\sin }^{-1}}\left( -0.4472 \right)$ We have to calculate the equation with the help of a calculator in radian mode: $\theta ={{\sin }^{-1}}\left( +0.4472 \right)\approx 0.4636$ $\theta ={{\sin }^{-1}}\left( -0.4472 \right)\approx -0.4636$ So, $x\approx 0.4636$ Also, the sine value is positive in the II quadrant and the value of $\pi $ is $3.14159$. Therefore, $\begin{align} & x=\pi -0.4636 \\ & =3.14159-0.4636 \\ & \approx 2.6780 \end{align}$ The sine value is negative in the III and IV quadrants and the value of $\pi $ is $3.14159$. So, $\begin{align} & x=\pi +0.4636 \\ & =3.14159+0.4636 \\ & \approx 3.6052 \end{align}$ Or, $\begin{align} & x=2\pi -0.4636 \\ & =2\left( 3.14159 \right)-0.4636 \\ & =6.28318-0.4636 \\ & \approx 5.8195 \end{align}$ Thus, the solution on the interval $[0,2\pi )$, correct to four decimal places is $0.4636$ , $2.6780$ , $3.6052$ or $5.8195$.
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