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 326: 47

Answer

$$A= n360^{\circ},\ \ \text{or}\ \ 120^{\circ}+n360^{\circ} $$ where $n $ is integer

Work Step by Step

Given $$ \sin (A+30^{\circ}) =\left( \frac{1}{2}\right)$$ Since \begin{align*} A+30^{\circ}&=\sin^{-1}\left( \frac{1}{2}\right)\\ A+30^{\circ}&=30^{\circ},\ \ \text{or}\ \ 150^{\circ}\\ A&= 0^{\circ},\ \ \text{or}\ \ 120^{\circ} \end{align*} Since the period of $\sin$ function is $360^{\circ}$ , then the general solution is $$A= n360^{\circ},\ \ \text{or}\ \ 120^{\circ}+n360^{\circ} $$ where $n $ is integer
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