We are asked to solve the equation $\sin(2x) + \sin(x) = 0$ in the interval $[0, 2\pi)$.

Step 1: Use the double-angle identity for sine

We can rewrite $\sin(2x)$ using the identity: $\sin(2x) = 2\sin(x)\cos(x)$ Substituting this into the equation, we get: $2\sin(x)\cos(x) + \sin(x) = 0$

Step 2: Factor the equation

Now, factor out $\sin(x)$ from both terms: $\sin(x)(2\cos(x) + 1) = 0$

Step 3: Solve for $x$

We have two cases to consider:

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

The solutions to $\sin(x) = 0$ in the interval $[0, 2\pi)$ are: $x = 0, \pi$

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

Solve for $\cos(x)$: $2\cos(x) = -1 \quad \Rightarrow \quad \cos(x) = -\frac{1}{2}$ The solutions to $\cos(x) = -\frac{1}{2}$ in the interval $[0, 2\pi)$ are: $x = \frac{2\pi}{3}, \frac{4\pi}{3}$

Final solutions:

Combining all solutions from both cases, we have: $x = 0, \pi, \frac{2\pi}{3}, \frac{4\pi}{3}$

These are the solutions in the interval $[0, 2\pi)$.

Would you like more details on any part of the solution? Here are some related questions for further exploration:

1. How do you derive the double-angle identity for sine?
2. What are the fundamental identities for trigonometric functions?
3. How do you solve trigonometric equations with multiple solutions in a given interval?
4. How do you solve equations involving both sine and cosine?
5. How do you graph trigonometric functions to visualize solutions?

Tip: Always check the interval given in the problem and make sure your solutions are within that range before finalizing your answer.