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 - Concept and Vocabulary Check - Page 703: 2


The solution of $\tan x=-\sqrt{3}$ in $[0,\pi )$ is $x=\pi -\frac{\pi }{3}$, or $x=\underline{\frac{2\pi }{3}}$. If n is any integer, all solutions of $\tan x=-\sqrt{3}$ are given by $\frac{2\pi }{3}+n\pi $.

Work Step by Step

To calculate another value of x: $\begin{align} & x=\pi -\frac{\pi }{3} \\ & =\frac{3\pi -\pi }{3} \\ & =\frac{2\pi }{3} \end{align}$ Therefore, the value of x will be $\frac{2\pi }{3}$. The interval $[0,\pi )$ covers the 1st and 2nd quadrants. The value of x is $\frac{2\pi }{3}$, according to the standard trigonometric general solution. Since, all the solutions for $\tan x=-\sqrt{3}$ can be written as $\frac{2\pi }{3}+n\pi $. The value is considered to be $\pi $ because the interval has mentioned the limit of $[0,\pi )$. Thus, the values of x will be $\frac{2\pi }{3}\,\text{ and }\,\frac{2\pi }{3}+n\pi $.
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