<s>[INST] Range of $\sin^2x-\sin x +1$ How can we find the range of $f(x) =\sin^2x-\sin x +1$? 
The function can be written as $(\sin x-\frac{1}{2})^2+\frac{3}{4}$.
Range of $\sin x$ function is given by : $-1 \leq \sin x \leq 1$.  Please guide how to get the result.
 [/INST] $-1 \leq sinx \leq 1 \Rightarrow -\frac{3}{2} \leq sinx-\frac{1}{2} \leq \frac{1}{2} \Rightarrow 0 \leq (sinx - \frac{1}{2})^2 \leq \frac{9}{4} \Rightarrow \frac{3}{4}\leq(sinx-\frac{1}{2})^2 + \frac{3}{4}\leq 3$
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