## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 5 - Section 5.5 - Trigonometric Equations - Exercise Set - Page 703: 21

#### Answer

The solutions of the equation are $\theta =\frac{\pi }{6}+2n\pi$ or $\theta =\frac{5\pi }{6}+2n\pi$, where n is any integer.

#### Work Step by Step

\begin{align} & 4sin\theta -1=2\sin \theta \\ & 4sin\theta -2sin\theta =1 \\ & 2\sin \theta =1 \\ & \sin \theta =\frac{1}{2} \end{align} $sin\frac{\pi }{6}=\frac{1}{2}$, and the solutions of $\sin \theta =\frac{1}{2}$ in [0,2 $\pi$ ) are: $\theta =\frac{\pi }{6}$ \begin{align} & \theta =\pi -\frac{\pi }{6} \\ & =\frac{6\pi }{6}-\frac{\pi }{6} \\ & =\frac{5\pi }{6} \end{align} The period of the sine function is $2\pi$, and the solutions are given by: $\theta =\frac{\pi }{6}+2n\pi$ or $\theta =\frac{5\pi }{6}+2n\pi$, where n is any integer.

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