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: 58


The solution set is $$\{-0.9669+\pi n, 1.2886+\pi n, n\in Z\}$$

Work Step by Step

$$\tan x(\tan x-2)=5$$ 1) Solve the equation $$\tan x(\tan x-2)=5$$ $$\tan^2 x-2\tan x-5=0$$ Consider the equation as a quadratic formula, with $a=1, b=-2, c=-5$ - Calculate $\Delta$: $\Delta=b^2-4ac=(-2)^2-4\times1\times(-5)=4+20=24$ - Find out $\tan x$: $$\tan x=\frac{-b\pm\sqrt\Delta}{2a}=\frac{2\pm\sqrt{24}}{2}=\frac{2\pm2\sqrt6}{2}=1\pm\sqrt6$$ - For $\tan x=1+\sqrt6$. $x$ would be $$x=\tan^{-1}(1+\sqrt6)$$ $$x\approx1.2886$$ (Be careful with degrees and radians. Here we need to use radians.) - For $\tan x=1-\sqrt6$. $x$ would be $$x=\tan^{-1}(1-\sqrt6)$$ $$x\approx-0.9669$$ Therefore, overall, $$x=\{-0.9669,1.2886\}$$ 2) Solve the equation for all solutions - The integer multiples of the period of the tangent function is $\pi$. - We apply it to each solution found in step 1. The results are the solution set. So the solution set is $$\{-0.9669+\pi n, 1.2886+\pi n, n\in Z\}$$
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