Answer
$x=\dfrac{\pi}{6}, \dfrac{5\pi}{6}$
Work Step by Step
We have $(\sin x \cos \dfrac{\pi}{6}+\cos x \sin \dfrac{\pi}{6})-(\sin x \cos \dfrac{7\pi}{6}-\cos x \sin \dfrac{7\pi}{6}) =\dfrac{\sqrt 3}{2}$
$ (\dfrac{\sqrt 3}{2} \sin x +\dfrac{1}{2} \cos x)-(-\dfrac{\sqrt 3}{2} \sin x -(-\dfrac{1}{2}) \cos x)=\dfrac{\sqrt 3}{2} $
After simplifying, we get $\dfrac{\sqrt 3}{2} \sin x +\dfrac{1}{2} \cos x + \dfrac{\sqrt 3}{2} \sin x -\dfrac{1}{2} \cos x=\dfrac{\sqrt 3}{2}$
or, $ \sqrt 3 \sin x =\dfrac{\sqrt 3}{2}$
or, $\sin x=\dfrac{1}{2}$
This yields: $x=\dfrac{\pi}{6}, \dfrac{5\pi}{6}$
