Let's solve the given equation:

Problem:

Solve $\sin(2x) = \sin(x)$ for $0 \leq x < 2\pi$.

Step 1: Using Trigonometric Identities

The equation can be rewritten using the double-angle identity: $\sin(2x) = 2\sin(x)\cos(x)$ Substitute this into the equation: $2\sin(x)\cos(x) = \sin(x)$

Step 2: Factorize

Rearrange to factor out $\sin(x)$: $\sin(x)\left(2\cos(x) - 1\right) = 0$

This gives two cases:

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

Step 3: Solve Each Case

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

$\sin(x) = 0 \implies x = 0, \pi, 2\pi$ Since the interval is $0 \leq x < 2\pi$, we have: $x = 0, \pi$

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

Solve for $\cos(x)$: $\cos(x) = \frac{1}{2}$ This occurs at: $x = \frac{\pi}{3}, \frac{5\pi}{3}$

Step 4: Combine Solutions

The solutions for $x$ in the interval $[0, 2\pi)$ are: $x = 0, \pi, \frac{\pi}{3}, \frac{5\pi}{3}$

Let me know if you'd like further clarifications or a detailed graph! Here's how you can expand this understanding:

1. How does factoring help simplify trigonometric equations?
2. What are the key trigonometric identities needed for solving equations like this?
3. Can you use a unit circle to verify these solutions graphically?
4. Why do we exclude $x = 2\pi$ in this interval?
5. What would change if the interval were $0 \leq x \leq 2\pi$?

Tip: Always verify your solutions within the given interval by substituting them back into the original equation.