If $x\ne \frac{n\pi }{2}$and ${\left(\cos x\right)}^{{\sin }^{2}x-3\sin x+2}=1$ then the general solutions of x is

As $x\ne \frac{n\pi }{2}$$\Rightarrow \cos x\ne 0,1-1$

So,  ${\left(\cos x\right)}^{{\sin }^{2}x-3\sin x+2}=1$

$\Rightarrow {\sin }^{2}x-3\sin x+2=0$

$\Rightarrow \left(\sin x-2\right)\left(\sin x-1\right)=0$

$\Rightarrow \sin x=1,2$
Where  $\sin x=2$  is not possible, and $\sin x=1\Rightarrow x=2n\pi +\frac{\pi }{2}$  does not satisfy the given condition.