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: 100


The solution is $\underline{x=\frac{\pi }{2},\frac{7\pi }{6},\frac{3\pi }{2},\frac{11\pi }{6}}$.

Work Step by Step

We have to solve the equation: $\begin{align} & \sin 2x+\cos x=0 \\ & 2\sin x\cos x+\cos x=0 \end{align}$ And use the formula, $\sin 2\theta =2\sin \theta \cos \theta $ Factor: $co\operatorname{s}x\left( 2\sin x+1 \right)=0$ From the above result it can be concluded that, $\cos x=0$ Another term will be: $2\sin x+1=0$ Compute the value of x: $\begin{align} & \cos x=0 \\ & x=\left( 2n+1 \right)\frac{\pi }{2} \end{align}$ And the value of $\sin x$ will be: $\begin{align} & 2\sin x+1=0 \\ & 2\sin x=-1 \\ & \sin x=\frac{-1}{2} \end{align}$ So, in the interval $\left[ 0,2\pi \right),$ the sine function is $\frac{-1}{2}$ at $\frac{7\pi }{6}\text{ and }\frac{11\pi }{6}$, according to the trigonometric table. $\begin{align} & \sin x=\frac{-1}{2} \\ & x=\frac{7\pi }{6} \end{align}$ And another value is: $\begin{align} & \sin x=\frac{-1}{2} \\ & x=\frac{11\pi }{6} \end{align}$ The period of the sine function is $2\pi $; the general solution of the equation is: $x=\frac{7\pi }{6}+2n\pi $ and $x=\frac{11\pi }{6}+2n\pi $ To get different solutions, $\text{put }n=0,1,2,3\ldots $ $\begin{align} & \text{For }n=0, \\ & x=\left( 2n+1 \right)\frac{\pi }{2} \\ & =\frac{\pi }{2} \end{align}$ And the other value when n is 0: $\begin{align} & x=\frac{7\pi }{6}+2n\pi \\ & =\frac{7\pi }{6} \end{align}$ Or $\begin{align} & x=\frac{11\pi }{6}+2n\pi \\ & =\frac{11\pi }{6} \end{align}$ When the value of n is 1: $\begin{align} & \text{ }n=1 \\ & x=\left( 2n+1 \right)\frac{\pi }{2} \\ & =\frac{3\pi }{2} \end{align}$ When we put other values of $n$, the solution becomes outside the interval $\left[ 0,2\pi \right).$
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