Giải và biện luận:       $x^{\sin x - a}> 1     (1) $ với $0 < x < \frac{\pi }{2}$
*) $x = 1$  bất phương trình không thỏa mãn do tương đương $1>1$ (vô lý).
*) $x \ne 1$  ta xét các trường hợp sau:
$\begin{array}{l}
a)\,\,\,0 < x < 1:\\\,(1)\,\, \Leftrightarrow \left\{ \begin{array}{l}
\sin \,x - a < 0\\
0 < x < 1
\end{array} \right.\\
\,\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \left\{ \begin{array}{l}
\sin \,x < a\\
0 < x < 1
\end{array} \right.\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \left\{ \begin{array}{l}
x < \arcsin \,a\\
0 < x < 1
\end{array} \right.
\end{array}$
Nếu $a \le 0:$ bất phương trình vô nghiệm
         $0 < a < \sin \,1$ nghiệm của bất phương trình là : $0 < x < arc\sin \,\,a$
         $a \ge \sin \,1:$ nghiệm của bất phương trình là :$0 < x < 1$

$b)\,1 < x < \frac{\pi }{2}:$
      ${x^{{\mathop{\rm s}\nolimits} in\,x - a}} > 1\,\,\,\,\,\,\,\,\, \Leftrightarrow {\mathop{\rm s}\nolimits} in\,x - a > 0$
Ta có hệ:$\left\{ \begin{array}{l}
1 < x < \frac{\pi }{2}\\
\sin \,x > a
\end{array} \right.\,\,\,\,\,\,\,\,\,\,\Leftrightarrow \left\{ \begin{array}{l}
1 < x < \frac{\pi }{2}\\
x > arc\sin \,a
\end{array} \right.$
Nếu $a \le \arcsin \,1:$ $(1)$ có nghiệm là :$1 < x < \frac{\pi }{2}$
        $arc\sin \,1 < a < 1:$ nghiệm là $arcsin \, < x < \frac{\pi }{2}$
        $a \ge 1$:            $(1)$ vô nghiệm

Kết luận:
$1)$  $a \le 0:$nghiệm là $0 < x < \frac{\pi }{2}$
$2)$ $0 < a \le \arcsin \,1:$$\left[ \begin{array}{l}
0 < x < arc\sin \,\,a\\
0 < x < \pi
\end{array} \right.$
$3)$ $arc\sin \,1 < a < 1:$$\left[ \begin{array}{l}
0 < x < 1\\
arc\sin \, < x < \frac{\pi }{2}
\end{array} \right.$
$4)$ $a \ge 1$$0 < x < 1$

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