Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 6 - Inverse Circular Functions and Trigonometric Equations - Section 6.2 Trigonometric Equations I - 6.2 Exercises - Page 267: 58

Answer

The solution set is $$\{33.7^\circ+180^\circ n, 135^\circ+180^\circ n, n\in Z\}$$

Work Step by Step

$$\frac{2\cot^2\theta}{\cot\theta+3}=1$$ 1) Solve the equation over the interval $[0^\circ,360^\circ)$ $$\frac{2\cot^2\theta}{\cot\theta+3}=1$$ $$2\cot^2\theta=\cot\theta+3$$ $$2\cot^2\theta-\cot\theta-3=0$$ $$(2\cot^2\theta+2\cot\theta)+(-3\cot\theta-3)=0$$ $$2\cot\theta(\cot\theta+1)-3(\cot\theta+1)=0$$ $$(\cot\theta+1)(2\cot\theta-3)=0$$ $$\cot\theta=-1\hspace{1cm}\text{or}\hspace{1cm}\cot\theta=\frac{3}{2}$$ - For $\cot\theta=-1$ Over the interval $[0^\circ, 360^\circ)$, there are 2 values of $\theta$ where $\cot\theta=-1$, which are $135^\circ$ and $315^\circ$. So, $\theta=\{135^\circ, 315^\circ\}$ - For $\cot\theta=\frac{3}{2}$. That means, $$\theta=\cot^{-1}\frac{3}{2}$$ $$\theta=\tan^{-1}\frac{2}{3}$$ $$\theta\approx33.7^\circ\hspace{1cm}\text{or}\hspace{1cm}\theta\approx213.7^\circ$$ Therefore, overall, $$\theta=\{33.7^\circ,135^\circ, 213.7^\circ, 315^\circ\}$$ 2) Solve the equation for all solutions - The integer multiples of the period of the cotangent function is $180^\circ$. - We apply it to each solution found in step 1. - We end up with a set like this $$\{33.7^\circ+180^\circ n, 135^\circ+180^\circ n, 213.7^\circ+180^\circ n, 315^\circ+180^\circ n,nāˆˆZ\}$$ - However, as $33.7^\circ+180^\circ n$ and $213.7^\circ+180^\circ n$ refer to the same set of values, we only need to include one of them in the final solution set. The same thing can be said about $135^\circ+180^\circ n$ and $315^\circ+180^\circ n$. So the solution set is $$\{33.7^\circ+180^\circ n, 135^\circ+180^\circ n, n\in Z\}$$
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