Answer :
It is given that
[tex]\begin{gathered} 2\sin ^2x-\sin x=0 \\ \sin x(2\sin x-1)=0 \\ \sin x=0,2\sin x-1=0 \\ \sin x=0,\sin x=\frac{1}{2} \end{gathered}[/tex]Therefore in the interval
[tex]\lbrack0,2\pi)[/tex]sin function takes the value 0 at
[tex]x=0,x=\pi[/tex]It takes the value of 0.5 at
[tex]x=\frac{\pi}{6},\frac{5\pi}{6}[/tex]Therefore the above four values are correct
Option C is correct.