Trigonometry (11th Edition) Clone

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

Chapter 6 - Quiz (Sections 6.1-6.3) - Page 282: 5


The solution set is $$\{60^\circ, 120^\circ\}$$

Work Step by Step

$$2\sin\theta-\sqrt3=0$$ over the interval $[0^\circ,360^\circ)$ We carry out normal algebra here: $$\sin\theta=\frac{\sqrt3}{2}$$ $$\theta=\sin^{-1}\frac{\sqrt3}{2}$$ Over the interval $[0^\circ,360^\circ)$, there are two values of $\theta$ whose $\sin\theta=\frac{\sqrt3}{2}$, which are $60^\circ$ and $120^\circ$. In other words, $$\theta=\{60^\circ, 120^\circ\}$$ That is also the solution set to the problem.
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