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: 46

Answer

The solution set is $$\{\frac{5\pi}{6}+\pi n, n\in Z\}$$

Work Step by Step

$$\cot x+\sqrt3=0$$ 1) Solve the equation over the interval $[0,2\pi)$ $$\cot x+\sqrt3=0$$ $$\cot x=-\sqrt3$$ Over the interval $[0,2\pi)$, there are two values of $x$ where $\cot x=-\sqrt3$, which are $\frac{5\pi}{6}$ and $\frac{11\pi}{6}$. 2) Solve the equation for all solutions - The integer multiples of the period of the cotangent function is $\pi$. - We apply it to each solution found in step 1. The results are the solution set, which looks something like this $$\{\frac{5\pi}{6}+\pi n, \frac{11\pi}{6}+\pi n, n\in Z\}$$ - However, a closer look shows that both $\frac{5\pi}{6}+\pi n$ and $\frac{11\pi}{6}+\pi n$ refer to the same set of points, meaning in the final solution set, only one set needs to be included. Therefore, the final solution set is $$\{\frac{5\pi}{6}+\pi n, n\in Z\}$$
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