To solve the equation $\tan(x) + \sqrt{3} = 0$ in the interval $(0, 2\pi)$, follow these steps:

Step 1: Isolate $\tan(x)$

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

Step 2: Solve for $x$

The tangent function equals $-\sqrt{3}$ at angles where the reference angle is $\frac{\pi}{3}$, but in the quadrants where tangent is negative, which are the second and fourth quadrants.

Thus: $x = \pi - \frac{\pi}{3} \quad \text{(Second quadrant)}$ $x = 2\pi - \frac{\pi}{3} \quad \text{(Fourth quadrant)}$

Step 3: Simplify the results

$x = \frac{2\pi}{3}, \quad x = \frac{5\pi}{3}$

Final Answer:

$x = \frac{2\pi}{3}, \frac{5\pi}{3}$

Would you like more detailed steps or clarifications?

Related Questions:

1. How do we determine the reference angle for tangent equations?
2. Why is tangent negative in the second and fourth quadrants?
3. What are other examples of solving trigonometric equations like this one?
4. How do we generalize this solution beyond $(0, 2\pi)$?
5. Can this problem be solved graphically?

Tip:

When solving trigonometric equations, always analyze the signs of the function in the given interval to identify the correct quadrants.