We need to solve the trigonometric equation $\cos(\theta)(2\sin(\theta) - 1) = 0$ for $\theta$.

1. Problem Description

We need to solve the trigonometric equation $\cos(\theta)(2\sin(\theta) - 1) = 0$ for $\theta$.

2. Solution Steps

The given equation is $\cos(\theta)(2\sin(\theta) - 1) = 0$. For the product of two factors to be zero, at least one of the factors must be zero.

Thus, we have two cases:

Case 1: $\cos(\theta) = 0$

The solutions for $\cos(\theta) = 0$ are $\theta = \frac{\pi}{2} + n\pi$, where $n$ is an integer.

For $n = 0$, $\theta = \frac{\pi}{2}$.

For $n = 1$, $\theta = \frac{\pi}{2} + \pi = \frac{3\pi}{2}$.

Case 2: $2\sin(\theta) - 1 = 0$

This simplifies to $\sin(\theta) = \frac{1}{2}$.

The principal solution is $\theta = \frac{\pi}{6}$.

The other solution in the interval $[0, 2\pi)$ is $\theta = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$.

The general solutions are $\theta = \frac{\pi}{6} + 2n\pi$ and $\theta = \frac{5\pi}{6} + 2n\pi$, where $n$ is an integer.

Combining the solutions from both cases in the interval $[0, 2\pi)$:

$\theta = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{\pi}{6}, \frac{5\pi}{6}$.

3. Final Answer

$\theta = \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}, \frac{3\pi}{2}$