To solve the equation $\sin(2x) = \sin(x)$ within the interval $0 \leq x \leq 2\pi$, we can use trigonometric identities and properties of sine functions.

Step 1: Use the identity for $\sin(2x)$

We know that: $\sin(2x) = 2\sin(x)\cos(x)$ So, the equation becomes: $2\sin(x)\cos(x) = \sin(x)$

Step 2: Factor the equation

We can factor out $\sin(x)$ from both sides: $\sin(x) \cdot (2\cos(x) - 1) = 0$

This equation will be satisfied if either: $\sin(x) = 0$ or $2\cos(x) - 1 = 0$

Step 3: Solve each equation separately

Case 1: $\sin(x) = 0$

The sine function equals zero at: $x = 0, \pi, 2\pi$

Case 2: $2\cos(x) - 1 = 0$

Solve for $\cos(x)$: $\cos(x) = \frac{1}{2}$ The cosine function equals $\frac{1}{2}$ at: $x = \frac{\pi}{3}, \frac{5\pi}{3}$

Step 4: Combine all solutions

The values of $x$ that satisfy the original equation are: $x = 0, \frac{\pi}{3}, \pi, \frac{5\pi}{3}, 2\pi$

Thus, the solution set is: $\boxed{0, \frac{\pi}{3}, \pi, \frac{5\pi}{3}, 2\pi}$

Would you like further details on how these solutions were found or have any other questions?

Here are five related questions you might find interesting:

1. How do we solve trigonometric equations involving tangent?
2. What are the general solutions for the equation $\sin(x) = k$?
3. How does the double-angle identity for sine function work in other equations?
4. How can we solve $\cos(2x) = \cos(x)$ for $0 \leq x \leq 2\pi$?
5. What are the zeros of the sine function on a given interval?

Tip: When solving trigonometric equations, always check if you can factor the equation to make it simpler to solve.