Answer
$$\theta= n\pi \hspace{4mm} \forall n\in Z $$ are the critical numbers of $f$.
Work Step by Step
How to find the critical numbers of a function $f$ according to the definition
1) Find all numbers of $c$ satisfying either $f'(c)=0$ or $f'(c)$ does not exist.
2) See that all the $c$ we get lie in the domain of $f$ $(D_f)$ or not.
- If $c$ lies in $D_f$, $c$ is a critical number of $f$.
- If $c$ does not lie in $D_f$, $c$ is not a critical number of $f$.
$$f(\theta)=2\cos\theta+\sin^2\theta$$
$D_f=R$
1) Now, we find $f'(\theta)$
$$f'(\theta)=-2\sin\theta+2\sin\theta(\sin\theta)'$$
$$f'(\theta)=-2\sin\theta+2\sin\theta\cos\theta$$
$$f'(\theta)=-2\sin\theta(1-\cos\theta)$$
We find $f'(\theta)=0$, $$f'(\theta)=0$$
$$-2\sin\theta(1-\cos\theta)=0$$
$$\sin\theta=0\hspace{0.5cm}or\hspace{0.5cm}\cos\theta=1$$
$$\theta=\pm n\pi\hspace{0.5cm}or\hspace{0.5cm}\theta=\pm2n\pi$$
which means, $$\theta= n\pi \hspace{4mm} \forall n\in Z $$
These values belong to $D_f$, so they are critical number of $f$.
There are no values of $\theta$ that can make $f'(\theta)$ not exist.
We conclude that $$\theta= n\pi \hspace{4mm} \forall n\in Z $$ are the critical numbers of $f$.
