1.

Find the range of `f(x) =sin^2x-3sinx+2`.

Answer» `f(x)=sin^2x-3sinx+2`
`=(sinx-3//2)^2+2-9//4`
`=(sinx-3//2)^2-1//4`
Now, `-1lesinxle1`
or `-5//2lesinx-3//2le-1//2`
or `1/4le(sinx-3//2)^2le25//4`
or `0le(sinx-3//2)^2-1//4le6`


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