If sin x + cosec x = 2,

We have:

$\sin x+\operatorname{cosec} x=2$

$\Rightarrow \sin x+\frac{1}{\sin x}=2$

$\Rightarrow \frac{\sin ^{2} x+1}{\sin x}=2$

$\Rightarrow \sin ^{2} x+1=2 \sin x$

$\Rightarrow \sin ^{2} x+1-2 \sin x=0$

$\Rightarrow(\sin x-1)^{2}=0$

$\Rightarrow \sin x-1=0$

$\Rightarrow \sin x=1$

And, $\operatorname{cosec} x=\frac{1}{\sin x}=1$

$\therefore \sin ^{n} x+\operatorname{cosec}^{n} x=1^{n}+1^{n}$

$=1+1=2$