The problem asks to find the inclination angle of the line $x + \sqrt{3}y + 1 = 0$.

1. Problem Description

The problem asks to find the inclination angle of the line $x + \sqrt{3}y + 1 = 0$.

2. Solution Steps

First, rewrite the equation of the line in the slope-intercept form $y = mx + b$, where $m$ is the slope.

Given the equation $x + \sqrt{3}y + 1 = 0$, we can isolate $y$:

$\sqrt{3}y = -x - 1$

$y = -\frac{1}{\sqrt{3}}x - \frac{1}{\sqrt{3}}$

The slope $m$ of the line is $-\frac{1}{\sqrt{3}}$.

The inclination angle $\theta$ satisfies $m = \tan(\theta)$.

Therefore, $\tan(\theta) = -\frac{1}{\sqrt{3}}$.

Since the range of inclination angle $\theta$ is $[0, \pi)$, we need to find an angle $\theta$ in this range such that $\tan(\theta) = -\frac{1}{\sqrt{3}}$.

We know that $\tan(\frac{\pi}{6}) = \frac{1}{\sqrt{3}}$. Since $\tan(\theta)$ is negative, the angle $\theta$ must be in the second quadrant.

We can find the angle in the second quadrant by using the identity $\tan(\pi - x) = -\tan(x)$.

Thus, $\theta = \pi - \frac{\pi}{6} = \frac{6\pi}{6} - \frac{\pi}{6} = \frac{5\pi}{6}$.

3. Final Answer

The inclination angle is $\frac{5\pi}{6}$.