Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 6 - Test - Page 293: 2b



Work Step by Step

RECALL: $y=\arcsin{x} \longrightarrow \sin{y}=x$ where $y$ is in the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$. Note that: $\sin{(\frac{\pi}{3})} = \frac{\sqrt3}{2}$ Since sine is an dd function, then $\sin{(-\theta)} = -\sin{\theta}$. Thus, $\sin{(-\frac{\pi}{3})} = -\sin{(\frac{\pi}{3})}=-\frac{\sqrt3}{2}$ Therefore, the solution to the given equation is $-\frac{\pi}{3}$.
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