Precalculus (6th Edition) Blitzer

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

Chapter 6 - Section 6.4 - Graphs of Polar Equations - Exercise Set - Page 755: 97


The values of x are $\frac{\pi }{6},\frac{5\pi }{6},\,\text{ and }\,\frac{3\pi }{2}$.

Work Step by Step

$\cos 2x-\sin x=0,\,0\le x\,<2\pi $ …… (1) Substituting the value of $\cos 2x$ as $\cos 2x=1-2{{\sin }^{2}}x$ in equation (1) we get, $\begin{align} & 1-2{{\sin }^{2}}x-\sin x=0 \\ & -2{{\sin }^{2}}x-\sin x+1=0 \end{align}$ Multiplying the above equation by $-1$ we get $\begin{align} & -2{{\sin }^{2}}x-\sin x+1=0 \\ & 2{{\sin }^{2}}x+\sin x-1=0 \end{align}$ We can also express the equation as below: $\begin{align} & 2{{\sin }^{2}}x+\sin x-1=0 \\ & 2{{\sin }^{2}}x+2\sin x-\sin x-1=0 \end{align}$ Now, factorize the equation. $\begin{align} & 2\sin x\left( \sin x+1 \right)-1\left( \sin x+1 \right)=0 \\ & \left( 2\sin x-1 \right)\left( \sin x+1 \right)=0 \end{align}$ Solve for x. $2\sin x-1=0$ or $\sin x+1=0$ $2\sin x=1$ $\sin x=-1$ $\sin x=\frac{1}{2}$ $x=\frac{3\pi }{2}$ $x=\frac{\pi }{6}$ In the range $0\le x\,<2\pi $, the values of x for $2\sin x-1=0$ are $x=\pi -\frac{\pi }{6}=\frac{5\pi }{6}$.
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