<s>[INST] Find the critical numbers of the function. $$\sin^2(x) + \cos(x)$$
$$\{0 < x < 2\pi\}$$
I thought the answer would be $\pi$, but it is not. Can anyone explain why the answer is not $\pi$?
 [/INST] Taking the derivative:
$$\dfrac{d}{dx}\left(\sin^2 x + \cos x\right)=2\sin x \cos x -\sin x$$
We want $2\sin x\cos x-\sin x$ to equal $0$ (provided that $0 < x < 2\pi$). We immediately see that when $\sin x = 0$, $2\sin x \cos x-\sin x =0$. The only value of $x$ that will make $\sin x = 0$ and is in the interval $0<x<2\pi$ is $\pi$.
Now, when $\cos x=\frac{1}{2}$, $2\sin x \cos x-\sin x = \sin x-\sin x =0$. There are two values of $x$ that will make $\cos x =0$ and are in the interval. You should know that they are: $x=\frac{\pi}{3}$ and $x=\frac{5\pi}{3}$. Therefore your critical points are:
$$\displaystyle \color{green}{\boxed{x=\pi, \ \dfrac{\pi}{3}, \ \dfrac{5\pi}{3}}}$$
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