Trigonometry (11th Edition) Clone

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

Chapter 6 - Inverse Circular Functions and Trigonometric Equations - Section 6.3 Trigonometric Equations II - 6.3 Exercises - Page 279: 24


The solution set is $$\{15^\circ, 75^\circ, 195^\circ,255^\circ\}$$

Work Step by Step

$$2\sqrt3\sin2\theta=\sqrt3$$ over interval $[0^\circ,360^\circ)$ 1) Find corresponding interval for $2\theta$ The interval for $\theta$ is $[0^\circ,360^\circ)$, which can also be written as the inequality: $$0^\circ\le\theta\lt360^\circ$$ Therefore, for $2\theta$, the inequality would be $$0^\circ\le2\theta\lt720^\circ$$ Thus, the corresponding interval for $2\theta$ is $[0^\circ,720^\circ)$. 2) Now we examine the equation: $$2\sqrt3\sin2\theta=\sqrt3$$ $$\sin2\theta=\frac{\sqrt3}{2\sqrt3}=\frac{1}{2}$$ Over interval $[0^\circ,720^\circ)$, there are 4 values whose sine equals $\frac{1}{2}$, which are $\{30^\circ, 150^\circ, 390^\circ,510^\circ\}$ Therefore, $$2\theta=\{30^\circ, 150^\circ, 390^\circ,510^\circ\}$$ It follows that $$\theta=\{15^\circ, 75^\circ, 195^\circ,255^\circ\}$$ This is the solution set of the equation.
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