As `x ne (npi)/2 rArr cosx ne 0,1,-1`
So, `(cosx)^(sin^(2)x-3sinx+2) = 1 rArr sin^(2)x-3sinx+2=0`
`therefore (sinx-2)(sinx-1)=0 rArr sinx=1,2`
Where `sinx=2` is not possible and `sinx=2` is not possible and `sinx=1` which is also not possible as `x ne (npi)/(2)`
`therefore` no general solutions is possible. Ans.