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 703: 27

Answer

The solutions in the interval $[0,2\pi )$ are $\frac{5\pi }{24}$, $\frac{7\pi }{24}$, $\frac{17\pi }{24}$, $\frac{19\pi }{24}$, $\frac{29\pi }{24}$, $\frac{31\pi }{24}$, $\frac{41\pi }{24}$ and $\frac{43\pi }{24}$.

Work Step by Step

We know that the period of the cosine function is $2\pi $. In the interval there are two values at which the cosine function is $-\frac{\sqrt{3}}{2}$. One is $\frac{5\pi }{6}$ . The cosine is negative in quadrant III, thus the other value is: $\begin{align} & 2\pi -\frac{5\pi }{6}=\frac{12\pi -5\pi }{6} \\ & =\frac{7\pi }{6} \end{align}$ Therefore, all the solutions to $\cos 4x=-\frac{\sqrt{3}}{2}$ are given by: $\begin{align} & 4x=\frac{5\pi }{6}+2n\pi \\ & x=\frac{5\pi }{24}+\frac{n\pi }{2} \end{align}$ Or, $\begin{align} & 4x=\frac{7\pi }{6}+2n\pi \\ & x=\frac{7\pi }{24}+\frac{n\pi }{2} \end{align}$ Where, n is any integer. And the solutions in the interval $[0,2\pi )$ are obtained by letting $n=0$, $n=1$, $n=2$, and $n=3$. The equation is calculated by taking first $n$ as 0 and then as 1, 2, and 3. It can be further simplified as follows. $\begin{align} & x=\frac{5\pi }{24}+\frac{n\pi }{2} \\ & =\frac{5\pi }{24}+\frac{0\times \pi }{2} \\ & =\frac{5\pi }{24}+0 \\ & =\frac{5\pi }{24} \end{align}$ $\begin{align} & x=\frac{5\pi }{24}+\frac{n\pi }{2} \\ & =\frac{5\pi }{24}+\frac{1\times \pi }{2} \\ & =\frac{5\pi }{24}+\frac{1\pi }{2} \\ & =\frac{5\pi +12\pi }{24} \end{align}$ $=\frac{17\pi }{24}$ $\begin{align} & x=\frac{5\pi }{24}+\frac{n\pi }{2} \\ & =\frac{5\pi }{24}+\frac{2\times \pi }{2} \\ & =\frac{5\pi }{24}+\frac{2\pi }{2} \\ & =\frac{5\pi +24\pi }{24} \end{align}$ $=\frac{29\pi }{24}$ $\begin{align} & x=\frac{5\pi }{24}+\frac{n\pi }{2} \\ & =\frac{5\pi }{24}+\frac{3\times \pi }{2} \\ & =\frac{5\pi }{24}+\frac{3\pi }{2} \\ & =\frac{5\pi +36\pi }{24} \end{align}$ $=\frac{41\pi }{24}$ The second equation is also computed by taking first $n$ as 0 and then as 1, 2, and 3. It can be further simplified as follows. $\begin{align} & x=\frac{7\pi }{24}+\frac{n\pi }{2} \\ & =\frac{7\pi }{24}+\frac{0\times \pi }{2} \\ & =\frac{7\pi }{24}+0 \\ & =\frac{7\pi }{24} \end{align}$ $\begin{align} & x=\frac{7\pi }{24}+\frac{n\pi }{2} \\ & =\frac{7\pi }{24}+\frac{1\times \pi }{2} \\ & =\frac{7\pi }{24}+\frac{\pi }{2} \\ & =\frac{7\pi +12}{24} \end{align}$ $=\frac{19\pi }{24}$ $\begin{align} & x=\frac{7\pi }{24}+\frac{n\pi }{2} \\ & =\frac{7\pi }{24}+\frac{2\times \pi }{2} \\ & =\frac{7\pi }{24}+\frac{2\pi }{2} \\ & =\frac{7\pi +24}{24} \end{align}$ $=\frac{31\pi }{24}$ $\begin{align} & x=\frac{7\pi }{24}+\frac{n\pi }{2} \\ & =\frac{7\pi }{24}+\frac{3\times \pi }{2} \\ & =\frac{7\pi }{24}+\frac{3\pi }{2} \\ & =\frac{7\pi +36}{24} \end{align}$ $=\frac{43\pi }{24}$
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