Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 5 - Section 5.5 - Trigonometric Equations - Concept and Vocabulary Check - Page 703: 9


If $\sin 0.9695\approx 0.8246$, then the solutions of $\sin x=-0.8246$, $0\le x\le 2\pi $, are given by $x\approx \underline{2\pi }+0.9695$ and $x\approx \underline{2\pi }-0.9695$.

Work Step by Step

It is given in the problem that the values lie between $0\le x\le 2\pi $. And the values of the sine function are negative in the third and fourth quadrant. When 0.9695 is added to $2\pi $ the value remains negative and lies in the fourth quadrant and if $2\pi $ is subtracted from 0.9695, the value again remains negative. Thus, $x\approx 2\pi +0.9695$ and $x=2\pi -0.9695$.
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