To solve the equation:

$\tan(x) + \sqrt{3} = 0$

we can isolate $\tan(x)$:

$\tan(x) = -\sqrt{3}$

Now, we need to find the values of $x$ such that $\tan(x) = -\sqrt{3}$.

1. Identify the reference angle: $\tan(x) = \sqrt{3}$ at $x = \frac{\pi}{3}$ (or $60^\circ$) in radians.

2. Find the angles where $\tan(x) = -\sqrt{3}$: The tangent function is negative in the second and fourth quadrants. Thus, the solutions are:

   $x = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$ and $x = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}$

3. Include the periodicity of tangent: Since $\tan(x)$ has a period of $\pi$, the general solutions are:

   $x = \frac{2\pi}{3} + n\pi$ and $x = \frac{5\pi}{3} + n\pi$

where $n$ is any integer.

Therefore, the solutions to the equation are:

$x = \frac{2\pi}{3} + n\pi \quad \text{and} \quad x = \frac{5\pi}{3} + n\pi$

for any integer $n$.

Would you like more details on any part of this solution or have any questions?

Here are five related questions you might find useful:

1. How do we determine the reference angle for trigonometric functions?
2. What is the significance of the period in trigonometric solutions?
3. How does the sign of tangent affect the quadrant where the solution lies?
4. How would the solution differ if the equation was $\tan(x) - \sqrt{3} = 0$?
5. What are the solutions to other trigonometric equations, like $\sin(x) = -\frac{1}{2}$?

Tip: Always check which quadrants a trigonometric function's value is positive or negative to identify potential solutions accurately.