The problem involves analyzing the function $f(x) = 3\cos^2(x)$ within the interval $-\pi \leq x \leq \pi$ and determining its increasing and decreasing intervals. Here's how we solve it:

Step 1: Differentiate the function

The function is $f(x) = 3\cos^2(x)$. Using the chain rule:

$f'(x) = 3 \cdot 2\cos(x) \cdot (-\sin(x)) = -6\cos(x)\sin(x)$

Simplify further:

$f'(x) = -3\sin(2x) \quad \text{(using the double-angle formula: } \sin(2x) = 2\sin(x)\cos(x)\text{)}.$

Step 2: Critical points

To find critical points, solve $f'(x) = 0$:

$-3\sin(2x) = 0 \quad \Rightarrow \quad \sin(2x) = 0.$

The solutions for $\sin(2x) = 0$ in the interval $-\pi \leq x \leq \pi$ are:

$2x = n\pi, \quad n \in \mathbb{Z}.$

Dividing by 2, we get:

$x = n\frac{\pi}{2}.$

In $-\pi \leq x \leq \pi$, the specific solutions are:

$x = -\pi, -\frac{\pi}{2}, 0, \frac{\pi}{2}, \pi.$

Step 3: Test intervals

To determine where $f(x)$ is increasing or decreasing, analyze the sign of $f'(x) = -3\sin(2x)$ in each interval defined by the critical points:

• Interval $(-\pi, -\frac{\pi}{2})$: Pick $x = -\frac{3\pi}{4}$: $\sin(2x) = \sin(-\frac{3\pi}{2}) = 1 \quad \Rightarrow \quad f'(x) = -3(1) = -3 \quad (\text{decreasing}).$

• Interval $(-\frac{\pi}{2}, 0)$: Pick $x = -\frac{\pi}{4}$: $\sin(2x) = \sin(-\frac{\pi}{2}) = -1 \quad \Rightarrow \quad f'(x) = -3(-1) = 3 \quad (\text{increasing}).$

• Interval $(0, \frac{\pi}{2})$: Pick $x = \frac{\pi}{4}$: $\sin(2x) = \sin(\frac{\pi}{2}) = 1 \quad \Rightarrow \quad f'(x) = -3(1) = -3 \quad (\text{decreasing}).$

• Interval $(\frac{\pi}{2}, \pi)$: Pick $x = \frac{3\pi}{4}$: $\sin(2x) = \sin(\frac{3\pi}{2}) = -1 \quad \Rightarrow \quad f'(x) = -3(-1) = 3 \quad (\text{increasing}).$

Step 4: Conclusion

The function is:

• Decreasing on $(-\pi, -\frac{\pi}{2}) \cup (0, \frac{\pi}{2})$,
• Increasing on $(-\frac{\pi}{2}, 0) \cup (\frac{\pi}{2}, \pi)$.

Let me know if you want additional details or have questions!

Here are 5 related questions to expand understanding:

1. How does the double-angle formula help in simplifying trigonometric derivatives?
2. What are the implications of critical points in determining a function's behavior?
3. How can this process be generalized for other trigonometric functions like $\sin^2(x)$?
4. How would the analysis change if the interval were extended to $[-2\pi, 2\pi]$?
5. Why does the derivative test work for determining increasing and decreasing intervals?

Tip: Always simplify trigonometric derivatives using known identities to make calculations manageable!