Answer
(a)$x=\frac{7\pi}{6}+2n\pi,\ \ x= \frac{7\pi}{6}+2n\pi \ \text{or} x=\frac{\pi}{2} +2n\pi $
(b) $ \bigg\{\frac{\pi}{2} , \frac{7\pi}{6},\frac{11\pi}{6} \bigg\} $
Work Step by Step
(a) Given
$$2\sin^2 x-\sin x-1 =0 $$
Since
\begin{align*}
2\sin^2 x-\sin x-1&=0\\
(2\sin x+1)(\sin x-1)& =0
\end{align*}
Then
$$2\sin x+1=0,\ \ \text{or} \ \ \sin x-1=0$$
Case 1 , $$\sin x=\frac{-1}{2},\ \ \Rightarrow \ \ x=\frac{7\pi}{6}+2n\pi,\ \ \text{or}\ \ x= \frac{7\pi}{6}+2n\pi $$
Case 2 , $$\sin x =1,\ \ \Rightarrow \ \ x=\frac{\pi}{2} +2n\pi $$
Hence the general solutions is
$$x=\frac{7\pi}{6}+2n\pi,\ \ x= \frac{7\pi}{6}+2n\pi \ \text{or} x=\frac{\pi}{2} +2n\pi $$
(b) To find solutions on the interval $ 0\leq x<2\pi $, put values of $n $
$$ \bigg\{\frac{\pi}{2} , \frac{7\pi}{6},\frac{11\pi}{6} \bigg\} $$
