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 703: 23

Answer

The solutions of the equation are $\theta =\frac{3\pi }{2}+2n\pi $, where n is any integer.

Work Step by Step

$\begin{align} & 3sin\theta +5=-2\sin \theta \\ & 3sin\theta +2sin\theta =-5 \\ & 5\sin \theta =-5 \\ & \sin \theta =-1 \end{align}$ $sin\frac{\pi }{2}=1$, and the solutions of $\sin \theta =-1$ in [0,2 $\pi $ ) are: $\begin{align} & \theta =\pi +\frac{\pi }{2} \\ & =\frac{2\pi }{2}+\frac{\pi }{2} \\ & =\frac{3\pi }{2} \end{align}$ $\begin{align} & \theta =2\pi -\frac{\pi }{2} \\ & =\frac{4\pi }{2}-\frac{\pi }{2} \\ & =\frac{3\pi }{2} \end{align}$ The period of the sine function is $2\pi $, and the solutions are given by: $\theta =\frac{3\pi }{2}+2n\pi $, where n is any integer.
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