To find all values of $x$ such that $\sin(2x) = \sin(x)$ within the interval $0 \leq x \leq 2\pi$, we will proceed by solving the equation.

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 the equation as: $\sin(x) \cdot (2\cos(x) - 1) = 0$ This equation is satisfied when either:

1. $\sin(x) = 0$
2. $2\cos(x) - 1 = 0$

Step 3: Solve for $x$ from each factor

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

$\sin(x) = 0$ within $0 \leq x \leq 2\pi$ at: $x = 0, \pi, 2\pi$

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

Solve for $\cos(x)$: $2\cos(x) - 1 = 0 \implies \cos(x) = \frac{1}{2}$ $\cos(x) = \frac{1}{2}$ within $0 \leq x \leq 2\pi$ at: $x = \frac{\pi}{3}, \frac{5\pi}{3}$

Step 4: Combine all solutions

The values of $x$ that satisfy $\sin(2x) = \sin(x)$ within the interval $0 \leq x \leq 2\pi$ are: $x = 0, \frac{\pi}{3}, \pi, \frac{5\pi}{3}, 2\pi$

Would you like to go over any of these steps in more detail?

Here are some related questions you might find helpful:

1. How do you solve trigonometric equations involving multiple angles?
2. What are the general solutions for $\sin(x) = 0$ and $\cos(x) = \frac{1}{2}$?
3. How can you use trigonometric identities to simplify equations?
4. What other identities could be useful in solving trigonometric equations?
5. How can you verify the solutions of trigonometric equations within a specific interval?

Tip: When solving trigonometric equations, always consider the interval given and ensure that all solutions lie within that interval.