Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 5 - Review Exercises - Page 709: 66



Work Step by Step

Step 1. Rewrite the equation to get $1+tan^2x=4tan(x)-2$ or $tan^2x-4tan(x)+3=(tan(x)-1)(tan(x)-3)=0$ Step 2. For $tan(x)=1$, we have solutions in $[0,2\pi)$ as $x=\frac{\pi}{4}, \frac{5\pi}{4}$ Step 3. For $tan(x)=3$, we have solutions in $[0,2\pi)$ as $x=tan^{-1}3\approx1.2490$ and $x=\pi+1.2490\approx4.3906$ Step 4. Thus, we have all the solutions in $[0,2\pi)$ as $x=\frac{\pi}{4},1.2490,\frac{5\pi}{4},4.3906$
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