## Algebra 2 (1st Edition)

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