The problem is to solve the equation $\sin(\theta - \frac{\pi}{3}) = -\frac{1}{2}$ for $0 \le \theta < 2\pi$.

1. Problem Description

The problem is to solve the equation $\sin(\theta - \frac{\pi}{3}) = -\frac{1}{2}$ for $0 \le \theta < 2\pi$.

2. Solution Steps

Let $x = \theta - \frac{\pi}{3}$. The equation becomes $\sin(x) = -\frac{1}{2}$. We are given the condition $0 \le \theta < 2\pi$. This can be written as $-\frac{\pi}{3} \le \theta - \frac{\pi}{3} < 2\pi - \frac{\pi}{3}$, or $-\frac{\pi}{3} \le x < \frac{5\pi}{3}$.

We need to find the angles $x$ in the interval $[-\frac{\pi}{3}, \frac{5\pi}{3})$ such that $\sin(x) = -\frac{1}{2}$.

We know that $\sin(\frac{7\pi}{6}) = -\frac{1}{2}$ and $\sin(\frac{11\pi}{6}) = -\frac{1}{2}$. We want to check if these angles are in the specified interval.

Since $\frac{7\pi}{6} = \frac{7}{6}\pi \approx 1.167\pi$ and $\frac{11\pi}{6} = \frac{11}{6}\pi \approx 1.833\pi$, we have $0 < \frac{7\pi}{6} < \frac{5\pi}{3}$ and $0 < \frac{11\pi}{6} < \frac{5\pi}{3}$. Also, $-\frac{\pi}{3} < 0$, so $\frac{7\pi}{6}$ and $\frac{11\pi}{6}$ are in the given interval.

Now, $x = \theta - \frac{\pi}{3}$. Therefore, $\theta = x + \frac{\pi}{3}$.

When $x = \frac{7\pi}{6}$, $\theta = \frac{7\pi}{6} + \frac{\pi}{3} = \frac{7\pi}{6} + \frac{2\pi}{6} = \frac{9\pi}{6} = \frac{3\pi}{2}$.

When $x = \frac{11\pi}{6}$, $\theta = \frac{11\pi}{6} + \frac{\pi}{3} = \frac{11\pi}{6} + \frac{2\pi}{6} = \frac{13\pi}{6}$.

Since $0 \le \theta < 2\pi$, we have $0 \le \frac{3\pi}{2} < 2\pi$, so $\theta = \frac{3\pi}{2}$ is a solution. Since $0 \le \frac{13\pi}{6} = 2\pi + \frac{\pi}{6}$, we need to reduce it to an angle in the given interval, but $\frac{13\pi}{6}$ is outside the range. However, $\frac{13\pi}{6} = \frac{11\pi}{6} + \frac{2\pi}{6} = \frac{11\pi}{6} + \frac{\pi}{3}$, so $\theta = \frac{13\pi}{6}$ is another solution for $\theta$.

However, we must have $\theta < 2\pi$. Thus, $\theta = \frac{13\pi}{6} > 2\pi = \frac{12\pi}{6}$. Then, since $\theta$ must be less than $2\pi$, and the interval of $x$ is $[-\frac{\pi}{3}, \frac{5\pi}{3})$, and we know $\sin(\theta - \frac{\pi}{3}) = -\frac{1}{2}$, the solutions are $x = \frac{7\pi}{6}$ and $x = \frac{11\pi}{6}$. Hence $\theta = \frac{9\pi}{6} = \frac{3\pi}{2}$ and $\theta = \frac{13\pi}{6}$. But $\frac{13\pi}{6} > 2\pi$, so $\frac{13\pi}{6}$ is not a solution.

3. Final Answer

$\theta = \frac{3\pi}{2}$