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: 11


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

Work Step by Step

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