Determine Value of x for Triangle Side Lengths Given Angle

To find the value(s) of $x$, we apply the Law of Cosines to $\triangle \text{ABC}$ for the angle $\angle \text{ACB} = C = \frac{\pi}{6}$:

$\cos C = \frac{a^2 + b^2 - c^2}{2ab}$

Given the side lengths:
$a = x^2 + x + 1$
$b = x^2 - 1$
$c = 2x + 1$

First, we compute the numerator $a^2 + b^2 - c^2$:
$a^2 = (x^2 + x + 1)^2 = x^4 + 2x^3 + 3x^2 + 2x + 1$
$b^2 = (x^2 - 1)^2 = x^4 - 2x^2 + 1$
$c^2 = (2x + 1)^2 = 4x^2 + 4x + 1$

Adding $a^2$ and $b^2$, and subtracting $c^2$:
$a^2 + b^2 - c^2 = (2x^4 + 2x^3 + x^2 + 2x + 2) - (4x^2 + 4x + 1)$
$a^2 + b^2 - c^2 = 2x^4 + 2x^3 - 3x^2 - 2x + 1$

We can factor this polynomial by grouping or using the rational root theorem. Factoring out $(x^2 - 1)$:
$2x^4 + 2x^3 - 3x^2 - 2x + 1 = (x^2 - 1)(2x^2 + 2x - 1)$

Since $b = x^2 - 1$, we have:
$a^2 + b^2 - c^2 = b(2x^2 + 2x - 1)$

Substituting this into the cosine formula:
$\cos\left(\frac{\pi}{6}\right) = \frac{b(2x^2 + 2x - 1)}{2ab}$

Since the side length $b > 0$, we have $b \neq 0$, so we can cancel $b$:
$\frac{\sqrt{3}}{2} = \frac{2x^2 + 2x - 1}{2(x^2 + x + 1)}$

Multiplying both sides by $2(x^2 + x + 1)$:
$\sqrt{3}(x^2 + x + 1) = 2x^2 + 2x - 1$

Let $y = x^2 + x$. The equation becomes:
$\sqrt{3}(y + 1) = 2y - 1$
$\sqrt{3}y + \sqrt{3} = 2y - 1$
$(2 - \sqrt{3})y = \sqrt{3} + 1$
$y = \frac{\sqrt{3} + 1}{2 - \sqrt{3}} = (\sqrt{3} + 1)(2 + \sqrt{3}) = 5 + 3\sqrt{3}$

Substituting back $y = x^2 + x$:
$x^2 + x - (5 + 3\sqrt{3}) = 0$

Using the quadratic formula to solve for $x$:
$x = \frac{-1 \pm \sqrt{1 + 4(5 + 3\sqrt{3})}}{2} = \frac{-1 \pm \sqrt{21 + 12\sqrt{3}}}{2}$

Notice that:
$(3 + 2\sqrt{3})^2 = 3^2 + 2 \cdot 3 \cdot 2\sqrt{3} + (2\sqrt{3})^2 = 9 + 12\sqrt{3} + 12 = 21 + 12\sqrt{3}$

Thus, $\sqrt{21 + 12\sqrt{3}} = 3 + 2\sqrt{3}$.

This gives two possible values for $x$:

1. $x = \frac{-1 + (3 + 2\sqrt{3})}{2} = \frac{2 + 2\sqrt{3}}{2} = 1 + \sqrt{3}$
2. $x = \frac{-1 - (3 + 2\sqrt{3})}{2} = \frac{-4 - 2\sqrt{3}}{2} = -(2 + \sqrt{3})$

Since $a, b, c$ represent the lengths of the sides of a triangle, all sides must be strictly positive:

• For $x = -(2 + \sqrt{3})$, we have $c = 2x + 1 = 2(-(2 + \sqrt{3})) + 1 = -3 - 2\sqrt{3} < 0$, which is not possible.
• For $x = 1 + \sqrt{3}$, we have:
  $a = (1+\sqrt{3})^2 + (1+\sqrt{3}) + 1 = 6 + 3\sqrt{3} > 0$
  $b = (1+\sqrt{3})^2 - 1 = 3 + 2\sqrt{3} > 0$
  $c = 2(1+\sqrt{3}) + 1 = 3 + 2\sqrt{3} > 0$

All side lengths are positive and satisfy the triangle inequalities ($b+c = 6 + 4\sqrt{3} > a$).

Hence, the only valid value of $x$ is:
$x = 1 + \sqrt{3}$