Answer
(a) $ x=\frac{\pi}{2} +2n\pi, x=\frac{\pi}{6} +2n\pi,\ \text{or}\ \ x=\frac{5\pi}{6} +2n\pi$
(b) $ \bigg\{\frac{\pi}{2} ,\ \frac{\pi}{6} ,\ \frac{5\pi}{6}\bigg\} $
Work Step by Step
Given
$$( \sin x-1)(2\sin x-1) =0 $$
Then
$$\sin x-1=0,\ \ \text{or} \ \ 2\sin x-1=0$$
Case 1 , $$\sin x=1,\ \ \Rightarrow \ \ x=\frac{\pi}{2} +2n\pi $$
Case 1 , $$\sin x=\frac{1}{2},\ \ \Rightarrow \ \ x=\frac{\pi}{6} +2n\pi,\ \text{or}\ \ x=\frac{5\pi}{6} +2n\pi$$
Hence the general solutions is
$$ x=\frac{\pi}{2} +2n\pi, x=\frac{\pi}{6} +2n\pi,\ \text{or}\ \ x=\frac{5\pi}{6} +2n\pi$$
(b) To find solutions on the interval , put values of $n $
$$ \bigg\{\frac{\pi}{2} ,\ \frac{\pi}{6} ,\ \frac{5\pi}{6}\bigg\} $$
