The problem asks to consider the set $A = \{ x \in [0, 2\pi] \mid \sin^2(x) < \cos^2(x) \}$ and to express $A$ as a union of intervals.

Step 1: Simplifying the inequality

We start with the inequality: $\sin^2(x) < \cos^2(x)$ This can be rewritten as: $\sin^2(x) - \cos^2(x) < 0$ Using the identity $\sin^2(x) + \cos^2(x) = 1$, we can express this as: $\sin^2(x) = 1 - \cos^2(x)$ Thus, the inequality becomes: $1 - 2\cos^2(x) < 0$ Simplifying further: $2\cos^2(x) > 1 \quad \text{or} \quad \cos^2(x) > \frac{1}{2}$ Taking the square root: $|\cos(x)| > \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$

Step 2: Finding the solution in $[0, 2\pi]$

Now we need to find where $\cos(x)$ satisfies $|\cos(x)| > \frac{\sqrt{2}}{2}$. This inequality means that $\cos(x)$ must lie in the intervals where the cosine function is greater than $\frac{\sqrt{2}}{2}$ or less than $-\frac{\sqrt{2}}{2}$.

• $\cos(x) > \frac{\sqrt{2}}{2}$ occurs in the intervals $(0, \frac{\pi}{4})$ and $(7\pi/4, 2\pi)$.
• $\cos(x) < -\frac{\sqrt{2}}{2}$ occurs in the interval $(\frac{5\pi}{4}, \frac{7\pi}{4})$.

Step 3: Final solution

Therefore, the set $A$, where $\sin^2(x) < \cos^2(x)$, is expressed as the union of the intervals: $A = \left( 0, \frac{\pi}{4} \right) \cup \left( \frac{7\pi}{4}, 2\pi \right) \cup \left( \frac{5\pi}{4}, \frac{7\pi}{4} \right)$

Do you have any questions about this process?

Would you like additional details or further clarification?

Here are five related questions:

1. What is the general solution to $\cos(x) > \frac{\sqrt{2}}{2}$?
2. How do trigonometric identities help in simplifying inequalities?
3. What is the significance of $\cos^2(x) = \frac{1}{2}$ in trigonometric inequalities?
4. How does periodicity affect the solution of trigonometric inequalities?
5. Can you derive the same intervals for $\sin^2(x) > \cos^2(x)$?

Tip: Always check where the trigonometric functions satisfy the inequality within the given interval, accounting for periodic behavior.