The problem involves solving the compound inequality:

\[
\sin(2x) \leq \frac{1}{2} \quad \text{and} \quad \sin(2x) \geq \frac{1}{2}
\]

Let's break this down:

1. **Intersection of the Two Inequalities:**
   - The first inequality, \(\sin(2x) \leq \frac{1}{2}\), includes all values of \(x\) where the sine function is less than or equal to \(\frac{1}{2}\).
   - The second inequality, \(\sin(2x) \geq \frac{1}{2}\), includes all values of \(x\) where the sine function is greater than or equal to \(\frac{1}{2}\).

2. **Combine the Two Inequalities:**
   - The compound inequality \(\sin(2x) \leq \frac{1}{2} \) and \(\sin(2x) \geq \frac{1}{2} \) simplifies to:
     \[
     \sin(2x) = \frac{1}{2}
     \]

3. **Solve for \(x\):**
   - The general solution for \(\sin(2x) = \frac{1}{2}\) is:
     \[
     2x = n\pi + (-1)^n\frac{\pi}{6}
     \]
     where \(n\) is any integer.
   - Therefore, solving for \(x\):
     \[
     x = \frac{n\pi}{2} + \frac{(-1)^n\pi}{12}
     \]

This represents all the solutions \(x\) that satisfy both inequalities simultaneously.

Would you like further details or explanations on this solution?

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**Here are 5 related questions to consider:**

1. How does the sine function behave within different intervals of the unit circle?
2. What is the general solution to the equation \(\sin(2x) = 0\)?
3. How do you solve trigonometric inequalities involving cosine or tangent?
4. What are the specific values of \(x\) in the interval \([0, 2\pi)\) for which \(\sin(2x) = \frac{1}{2}\)?
5. How can the solution for \(\sin(2x) = \frac{1}{2}\) be verified graphically?

**Tip:** When solving trigonometric inequalities, always consider the periodic nature of the sine, cosine, and tangent functions to find the general solution.

Discover how to solve the trigonometric inequalities sin(2x) <= 1/2 and sin(2x) >= 1/2 with a detailed solution. This problem covers essential trigonometric concepts, including sine functions and their general solutions, and is suitable for students in Grades 10-12.

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