Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 6 - Review Exercises - Page 285: 44

Answer

$\left\{\frac{\pi}{6} +\pi(n), \frac{\pi}{3}+\pi(n)\text{where n is any integer}\right\}$

Work Step by Step

RECALL: $\sin{(2x)} = 2\sin{x}\cos{x}$ Thus, the given equation can be written as: $2(2\sin{x}\cos{x})=\sqrt3 \\2(\sin{(2x))}=\sqrt3 \\\sin{(2x)}=\frac{\sqrt3}{2}$ RECALL: $\sin{\theta}=a \longrightarrow \sin^{-1}{(a)}=\theta$ Thus, $\sin{(2x)} = \frac{\sqrt3}{2} \longrightarrow 2x=\sin^{-1}{(\frac{\sqrt3}{2})}$ Use a scientific calculator's inverse sine function to obtain $2x=\dfrac{\pi}{3} \text{ or } \dfrac{2\pi}{3} \\x=\dfrac{\frac{\pi}{3}}{2}\text{ or } \dfrac{\frac{2\pi}{3}}{2} \\x=\dfrac{\pi}{6} \text{ or } \dfrac{\pi}{3}$ The period of the function $\sin{(2x)}$ is $\pi$. This means that the function's value repeats every interval of $\pi$. Thus, if $\frac{\pi}{6}$ is a solution of the given equation, then $\frac{\pi}{6} + \pi(n)$, where $n$ is any integer, are all solutions of the equation. Similarly, if $\frac{\pi}{3}$ is a solution, then $\frac{\pi}{3}+\pi(n)$ where $n$ is any integer are all solutions of the equation. Thus, the solutions to the given equation are: $\left\{\frac{\pi}{6} +\pi(n), \frac{\pi}{3}+\pi(n)\text{where n is any integer}\right\}$
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