Answer
$\dfrac{\pi}{6}, \dfrac{5\pi}{6}$
Work Step by Step
We have $ \sin (x +\pi) -\sin x+1=1-2 \sin x=0$
After simplifying, we get $ \sin x \cos \pi +\cos x \sin \pi -\sin x+1=1-2 \sin x=0$
or, $\sin x-\sin x+1=1-2 \sin x=0$
or, $1= 2 \sin x$
or, $\sin x=\dfrac{1}{2}$
This yields: $ x=\dfrac{\pi}{6}, \dfrac{5\pi}{6}$
