We are given the trigonometric inequality:

$20 \sin(\theta) \geq 10 \quad \text{for} \quad 0 \leq \theta \leq \pi$

Step 1: Simplify the inequality

To solve for $\sin(\theta)$, divide both sides of the inequality by 20:

$\sin(\theta) \geq \frac{10}{20}$

Simplify the fraction:

$\sin(\theta) \geq \frac{1}{2}$

Step 2: Solve for $\theta$

We now need to find the values of $\theta$ where $\sin(\theta) \geq \frac{1}{2}$ in the interval $0 \leq \theta \leq \pi$.

We know that:

$\sin(\theta) = \frac{1}{2} \quad \text{at} \quad \theta = \frac{\pi}{6}$

Since $\sin(\theta)$ is increasing from $0$ to $\pi$, the inequality $\sin(\theta) \geq \frac{1}{2}$ holds for values of $\theta$ between $\frac{\pi}{6}$ and $\pi$.

Thus, the solution set is:

$\frac{\pi}{6} \leq \theta \leq \frac{\pi}{2}$

Step 3: Express the solution in interval notation

The solution in interval notation is:

$\left[\frac{\pi}{6}, \frac{\pi}{2}\right]$

---

Do you need any further details or clarification on this solution?

Here are 5 related questions to explore further:

1. How do you solve trigonometric inequalities in general?
2. What are the properties of the sine function over different intervals?
3. How can you solve similar inequalities involving other trigonometric functions like cosine or tangent?
4. What is the significance of the unit circle in solving trigonometric inequalities?
5. How do you handle trigonometric inequalities for other angles (greater than $\pi$)?

Tip: When solving trigonometric inequalities, it helps to sketch the graph of the trigonometric function within the given interval to visually identify the solution range.