$\begin{array}{l}
y = f\left( x \right) = {\sin ^2}x - 4\sin x + 8\\
t = \sin x\left( { - 1 \le t \le 1} \right)\\
\Rightarrow y = f\left( t \right) = {t^2} - 4t + 8\\
f\left( 1 \right) = 5,f\left( { - 1} \right) = 13\\
\to \max y = 13 \Rightarrow \sin x = - 1\\
\Rightarrow x = - \dfrac{\pi }{2} + k2\pi \left( {k \in \mathbb{Z}} \right)
\end{array}$
