$\begin{array}{l}
y = {\sin ^2}x - \sin x\\
t = \sin x\left( { - 1 \le t \le 1} \right)\\
\to f\left( t \right) = y = {t^2} - t,t \in \left[ { - 1;1} \right]\\
\Rightarrow y = \left( {{t^2} - t + \dfrac{1}{4}} \right) - \dfrac{1}{4} = {\left( {t - \dfrac{1}{2}} \right)^2} - \dfrac{1}{4} \ge - \dfrac{1}{4}\\
\Rightarrow \min y = - \dfrac{1}{4} \Rightarrow \sin x = \dfrac{1}{2} \Rightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{6} + k2\pi \\
x = \dfrac{{5\pi }}{6} + k2\pi
\end{array} \right.\left( {k \in \mathbb{Z}} \right)\\
y = f\left( t \right) = {t^2} - t\\
f\left( 1 \right) = 0,f\left( { - 1} \right) = 2 \Rightarrow \max y = 2\\
\Rightarrow t = - 1 \Leftrightarrow \sin x = - 1 \Leftrightarrow x = - \dfrac{\pi }{2} + k2\pi
\end{array}$
