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