How do I find the maximum and minimum of a sinusoidal function? I understand basic $\sin x$ and $\cos x$ min/max, but I am having a problem solving the minimum and maximum of the following:
$f(x) = \sin^2 x - \sin x$
Oh, and the range is $0 \le x \le \frac{3\pi}{2}$
 A: $$\begin{align*} f(x) &= \sin^2 x - \sin x \\ &= \sin^2 x - 2 \cdot \tfrac{1}{2} \sin x + (\tfrac{1}{2})^2 - \tfrac{1}{4} \\ &= \left( \sin x - \tfrac{1}{2} \right)^2 - \tfrac{1}{4}\end{align*}.$$  Because the square of a real number is nonnegative, $f$ attains a minimum if $\sin x = \frac{1}{2}$, and the consequences are straightforward.
To determine the maximum value, observe that $|\sin x| \le 1$; consequently, $f$ is maximized if $(\sin x - \tfrac{1}{2})^2$ is made as large as possible.  By inspection, this occurs if $\sin x = -1$.
A: $f'(x)=2\sin x \cos x-\cos x=\sin(2x)-\cos(x)=0$
$\implies \sin(2x)=\cos(x) \implies \cos(\frac{\pi}{2}-2x)=\cos(x) \implies x=2\pi n \pm \pi$.
Can you take it from there?
A: Just get rid of the $\sin$:
when  $x$ takes values in $[0,\frac{3\pi}2]$ $\sin$ takes all values of $[0,1]$.
We have to optimize $s^2 - s$ on this interval, this is on the top of the parabola: $$s = \frac 12$$
Going back to $\sin$, it takes the value $\frac 12$ one time on $[0,\frac{3\pi}2]$, when $$x = \frac\pi 6$$
