To find the critical numbers of the function $h(x) = \sin^2(x) + \cos(x)$ within the interval $0 < x < 2\pi$, we need to follow these steps:

1. Differentiate $h(x)$ with respect to $x$ to find $h'(x)$.

   $h(x) = \sin^2(x) + \cos(x)$

   Using the chain rule, the derivative of $\sin^2(x)$ is $2\sin(x) \cdot \cos(x)$ (which can also be written as $\sin(2x)$ due to the double-angle identity). Then, differentiating $\cos(x)$ gives $-\sin(x)$.

   So, $h'(x) = 2\sin(x)\cos(x) - \sin(x)$ $= \sin(x) (2\cos(x) - 1)$

2. Set $h'(x) = 0$ to find critical points:

   $\sin(x) (2\cos(x) - 1) = 0$

   This equation will be zero if either $\sin(x) = 0$ or $2\cos(x) - 1 = 0$.

   • For $\sin(x) = 0$: $x = 0, \pi, 2\pi$. However, since $0 < x < 2\pi$, we only consider $x = \pi$.

   • For $2\cos(x) - 1 = 0$: $\cos(x) = \frac{1}{2}$ This gives $x = \frac{\pi}{3}$ and $x = \frac{5\pi}{3}$ within the interval $0 < x < 2\pi$.

3. List the critical numbers in the interval $0 < x < 2\pi$:

   The critical numbers of $h(x)$ are: $x = \pi, \frac{\pi}{3}, \frac{5\pi}{3}$

Would you like a deeper explanation of any step or further analysis of these critical points?

Here are some related questions for further exploration:

1. How do we determine if these critical points are maximum, minimum, or saddle points?
2. Can we apply the second derivative test to confirm the nature of these critical points?
3. What are the values of $h(x)$ at these critical points?
4. How does the behavior of $\sin^2(x)$ affect the shape of $h(x)$ compared to just $\cos(x)$?
5. What is the impact of restricting the interval to $0 < x < 2\pi$ on the critical points?

Tip: For trigonometric functions, using identities like the double-angle formulas can simplify derivatives and make solving equations easier.