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 - Exercise Set - Page 704: 62

Answer

$\{ \frac{\pi}{2},\frac{3\pi}{2} \}$

Work Step by Step

Step 1. Factor the equation as $cot^2(x)(sin(x)-1)=0)$, which gives two solutions $cot(x)=0$ and $sin(x)=1$ Step 2. For $cot(x)=0$, we can find all x-values in $[0,2\pi)$ as $x=\frac{\pi}{2},\frac{3\pi}{2}$ Step 3. For $sin(x)=1$, we can find all x-values in $[0,2\pi)$ as $x=\frac{\pi}{2}$ Step 4. The solutions for the original equation in $[0,2\pi)$ are $\{ \frac{\pi}{2},\frac{3\pi}{2} \}$
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