To solve the equation $\sin(2x) - \sqrt{3} \cos(x) = 0$, we can use the double angle identity for sine:

$\sin(2x) = 2 \sin(x) \cos(x)$

Substituting this into the equation gives:

$2 \sin(x) \cos(x) - \sqrt{3} \cos(x) = 0$

Factoring out $\cos(x)$:

$\cos(x) (2 \sin(x) - \sqrt{3}) = 0$

This gives us two cases to consider:

1. $\cos(x) = 0$
2. $2 \sin(x) - \sqrt{3} = 0$

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

The solutions for $\cos(x) = 0$ are:

$x = \frac{\pi}{2} + n\pi, \quad n \in \mathbb{Z}$

Case 2: $2 \sin(x) - \sqrt{3} = 0$

Solving for $\sin(x)$:

$\sin(x) = \frac{\sqrt{3}}{2}$

The solutions for this are:

$x = \frac{\pi}{3} + 2k\pi \quad \text{and} \quad x = \frac{2\pi}{3} + 2k\pi, \quad k \in \mathbb{Z}$

Final Solutions

Combining both cases, the complete solution set is:

1. $x = \frac{\pi}{2} + n\pi$
2. $x = \frac{\pi}{3} + 2k\pi$
3. $x = \frac{2\pi}{3} + 2k\pi$

where $n, k \in \mathbb{Z}$.

Would you like more details on any specific part of this solution? Here are some related questions you might consider:

1. How do you graph the function to visualize the solutions?
2. What is the significance of the double angle identity in trigonometric equations?
3. Can you explain the unit circle's role in finding trigonometric solutions?
4. How can you verify these solutions using a calculator?
5. What would the solutions look like if the equation was changed to $\sin(2x) + \sqrt{3} \cos(x) = 0$?

Tip: Remember to check for extraneous solutions when dealing with trigonometric equations!