Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 7 - Trigonometric Identities and Equations - 7.6 Trigonometric Equations - 7.6 Exercises - Page 721: 39



Work Step by Step

Substitute t=$\sin\theta$: $ 9t^{2}-6t-1=0\qquad$ ... quadratic formula... $t=\displaystyle \frac{6\pm\sqrt{36+36}}{18}=\frac{6\pm 6\sqrt{2}}{18}$ $t=\displaystyle \frac{6(1\pm\sqrt{2})}{6(3)}$ Back-substitute $\displaystyle \sin\theta=\frac{1\pm\sqrt{2}}{3}$ In quadrant I, our calculator gives $\displaystyle \sin^{-1}(\frac{1+\sqrt{2}}{3})\approx$53.5849460516$\approx 53.6^{o}$ In q.II, $180^{0}-53.6=126.4^{o}$ For the other (negative) value, the calculator gives $\displaystyle \sin^{-1}(\frac{1-\sqrt{2}}{3})\approx$-7.93624951449$\approx-7.9^{o},$ which is not in the interval $[0,\ 360)$ , so we adjust with multiples of 180: (sines are negative in quadrants III and IV) In q. III, $180-(-7.9)=187.9^{o}$ In q.IV, $360^{o}-7.9=352.1^{o}$ Solution set = $\{53.6^{o},126.4^{o},187.9^{o},352.1^{o}\}$
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