<s>[INST] Absolute maximum value of $\sin^2(x)-\sin(x)$ in $[0,\frac{3\pi}{2}]$ I thought it does not have absolute maximum, but wanted to just check and see why
 [/INST] We are interested in $w^2-w$, with the proviso that $-1\le w\le 1$.
Completing the square, we find that 
$4w^2-4w=(2w-1)^2+3$.  So the minimum value of $4w^2-4w$ is $3$, reached when $w=1/2$. The maximum value in our interval is reached when $w$ is as negative as possible, namely at $w=-1$.
So the minimum value of $w^2-w$ is $\frac{3}{4}$, reached when $\sin x=\frac{1}{2}$, that is, at $x=\frac{\pi}{6}$ and also at $x=\pi-\frac{\pi}{6}$.
The maximum value of $w^2-w$ is $\frac{12}{4}$, reached when $\sin x=-1$, that is, at $x=\frac{3\pi}{2}$.
Remark: You came close to being right on no absolute maximum. If we replace the interval $[0,\frac{3\pi}{2}]$ by $(0,\frac{3\pi}{2})$, there is no longer an absolute maximum.
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