To solve the equation $\sin(x) = \sin\left(-\frac{\pi}{3}\right)$ in the interval $[0, 2\pi[$, follow these steps:

1. Simplify the equation: Using the periodic and symmetry properties of the sine function, we know: $\sin(-\theta) = -\sin(\theta).$ Hence: $\sin\left(-\frac{\pi}{3}\right) = -\sin\left(\frac{\pi}{3}\right) = -\frac{\sqrt{3}}{2}.$ Therefore, the equation becomes: $\sin(x) = -\frac{\sqrt{3}}{2}.$

2. Find the general solutions: The sine function is negative in the third and fourth quadrants. The reference angle associated with $-\frac{\sqrt{3}}{2}$ is: $\frac{\pi}{3}.$ Therefore, the solutions for $x$ in $[0, 2\pi[$ are: $x = \pi + \frac{\pi}{3} \quad \text{(third quadrant)},$ and $x = 2\pi - \frac{\pi}{3} \quad \text{(fourth quadrant)}.$

3. Simplify the solutions: $x = \pi + \frac{\pi}{3} = \frac{4\pi}{3},$ and $x = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}.$

4. Final solutions: The solutions to the equation in the interval $[0, 2\pi[$ are: $x = \frac{4\pi}{3} \quad \text{and} \quad x = \frac{5\pi}{3}.$

Would you like a detailed explanation of any of these steps or further help with similar problems?

Related Questions:

1. How do the symmetry properties of trigonometric functions help in solving equations?
2. What is the general solution for the equation $\sin(x) = k$ where $k$ is any constant?
3. How can the periodicity of sine and cosine simplify solving trigonometric equations?
4. How would the solutions change if the interval were different, e.g., $[-2\pi, 2\pi]$?
5. How can you use a unit circle to find solutions for trigonometric equations?

Tip: Always sketch a unit circle to visualize the angles where trigonometric functions take specific values!