Algebra 2 (1st Edition)

Published by McDougal Littell
ISBN 10: 0618595414
ISBN 13: 978-0-61859-541-9

Chapter 14 Trigonometric Graphs, Identities, and Equations - 14.6 Apply Sum and Difference Formulas - 14.6 Exercises - Skill Practice - Page 953: 35


$x=-arcsin \dfrac{1}{\sqrt 3}$ and $x=\pi+arcsin \dfrac{1}{\sqrt 3}$

Work Step by Step

Here, we have $\sin (x+\dfrac{5 \pi}{6})+ \sin (x -\dfrac{ 5\pi}{6})=1$ We know that $\sin a+\sin b=2 \sin (\dfrac{a+b}{2}) \cos (\dfrac{a-b}{2})$ $2 \sin x \cos (5 \pi/6)=1$ $ \implies 2 \sin x \times (\dfrac{-3}{2})=1$ This gives: $\sin x= \dfrac{1}{\sqrt 3}$ Thus, we have two solutions: $x=-arcsin \dfrac{1}{\sqrt 3}$ and $x=\pi+arcsin \dfrac{1}{\sqrt 3}$
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