Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 6 - Inverse Circular Functions and Trigonometric Equations - Section 6.2 Trigonometric Equations I - 6.2 Exercises - Page 273: 48


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

Work Step by Step

$$\tan\theta+1=0$$ 1) Solve the equation over the interval $[0^\circ,360^\circ)$ $$\tan\theta+1=0$$ $$\tan\theta=-1$$ Over the interval $[0^\circ, 360^\circ)$, there are two values of $\theta$ where $\tan\theta=-1$, which are $135^\circ$ and $315^\circ$. Therefore, $$\theta=\{135^\circ, 315^\circ\}$$ 2) Solve the equation for all solutions To find all solutions, we add the integer multiples of the period of tangent function, which is $180^\circ$, to each solution found in step 1. - There are two solutions found in step 1, $\{135^\circ, 315^\circ\}$, so it will be written as $\theta=135^\circ+180^\circ n$ and $\theta=315^\circ+180^\circ n$ where $n\in Z$. - However, a closer look reveals that $135^\circ+180^\circ n$ in fact covers all the possible values of $315^\circ+180^\circ n$, and vice versa. In other words, they are technically the same. So the solution set can be shortened to $$\{135^\circ+180^\circ n, n\in Z\}$$
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