Trigonometry 7th Edition

Published by Cengage Learning
ISBN 10: 1111826854
ISBN 13: 978-1-11182-685-7

Chapter 6 - Section 6.1 - Solving Trigonometric Equations - 6.1 Problem Set - Page 325: 11


$(a)$ $\displaystyle \theta=\frac{\pi}{3}+2k\pi $ or $\displaystyle \theta=\frac{2\pi}{3}+2k\pi$ $(b)$ $\displaystyle \frac{\pi}{3}$ and $\displaystyle \frac{2\pi}{3}.$

Work Step by Step

The first task is to isolate the trigonometric function on one side: Add $(\sqrt{3}-2\sin t)$ to both sides... $4\sin t-2\sin t=\sqrt{3}\qquad $ ... simplify $ 2\sin t=\sqrt{3}\quad$ ... divide with $2$ $\displaystyle \sin t=\frac{\sqrt{3}}{2}$ Now, we find a reference angle. From the table of characteristic angles, we know that $\displaystyle \sin\frac{\pi}{3}=\frac{\sqrt{3}}{2}.$ Next, we know that sine is positive in quadrants I and II, so angles (in radians) within the interval $0\leq t \lt 2\pi $ that satisfy the equation are $\displaystyle \frac{\pi}{3}$ and $\pi-\displaystyle \frac{\pi}{3}=\frac{2\pi}{3}$ Finally, to each individual solution, add multiples of $ 2\pi$ to cover all solutions: $(a)$ $\displaystyle \theta=\frac{\pi}{3}+2k\pi $ or $\displaystyle \theta=\frac{2\pi}{3}+2k\pi$ $(b)$ The solutions within the interval $0\leq t \lt 2\pi:$ $\displaystyle \frac{\pi}{3}$ and $\displaystyle \frac{2\pi}{3}.$
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