Trigonometry (10th Edition)

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

Chapter 6 - Test - Page 287: 16

Answer

$x=\dfrac{2\pi}{3}$

Work Step by Step

Divide $2\sqrt3$ to both sides of the equation: $\dfrac{2\sqrt3\sin{(\frac{x}{2})}}{2\sqrt3}=\dfrac{3}{2\sqrt3} \\\sin{(\frac{x}{2})}=\dfrac{\sqrt3}{2}$ RECALL: $\sin{x}=a \longrightarrow x=\sin^{-1}{(a)}$ Use the definition above to obtain: $\dfrac{\pi}{2}=\sin^{-1}{(\frac{\sqrt3}{2})}$ Use the inverse sine function of a calculator to obtain: \begin{array}{ccc} &\frac{x}{2} &= &\frac{\pi}{3} \\&2\cdot \frac{x}{2} &= &\frac{\pi}{3} \cdot 2 \\&x &= &\frac{2\pi}{3} \end{array}
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