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 $\pi$?
 A: The derivative is $2\sin x\cos x-\sin x$, which is $(\sin x)(2\cos x-1)$.
This is $0$ when $\sin x=0$, giving $x=\pi$,  and also when $\cos x=\frac{1}{2}$.  There are two values of $x$ in your interval where $\cos x=\frac{1}{2}$. 
A: 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:
$$\displaystyle \color{green}{\boxed{x=\pi, \ \dfrac{\pi}{3}, \ \dfrac{5\pi}{3}}}$$
A: Critical point is when the derivative is 0. The derivative of the function is:
$$2\sin(x)\cos(x) - \sin(x)=\sin(x)(2\cos(x)-1)$$
Setting this equal to $0$:
$$\sin(x)(2\cos(x)-1)=0$$
Remember your algebra rules~:
$$\sin(x)=0; 2\cos(x)-1=0$$
In the interval, $\sin(x)=0$ when $x=\pi$. $0$ and $2\pi$ are not included in the interval. So you are right about $\pi$.
Then for $2\cos(x)-1=0$:
$$\cos(x)=\frac{1}{2}$$
$x=\frac{\pi}{3}\text{ and } \frac{5\pi}{3}$
The critical points are then: $x=\pi, \frac{\pi}{3}, \frac{5\pi}{3}$.
