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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.

Consider the set A = { x ∈ [0, 2π] | sin^2(x) < cos^2(x) }. Express A as a union of intervals.

Explore how to solve the trigonometric inequality sin^2(x) < cos^2(x) for x in the interval [0, 2π]. This problem involves understanding trigonometric identities, inequalities, and expressing the solution as a union of intervals. Suitable for advanced high school students (Grades 11-12).

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