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.