Solve the equation $\frac{1}{\log_9 3} + \log_3(2x + 6) = \log_{\sqrt{3}}(1 - x) + 1$.

1. Evaluate the constant term involving logarithms.

$\frac{1}{\log_9 3} = \log_3 9 = 2$

The equation becomes:
$2 + \log_3(2x + 6) = \log_{\sqrt{3}}(1 - x) + 1$
$1 + \log_3(2x + 6) = \log_{\sqrt{3}}(1 - x)$

2. Change the base of the logarithm on the right-hand side to base 3.

$\log_{\sqrt{3}}(1 - x) = \frac{\log_3(1 - x)}{\log_3 \sqrt{3}} = \frac{\log_3(1 - x)}{\frac{1}{2}} = 2\log_3(1 - x)$

3. Combine the logarithmic terms into a single equation.

$1 + \log_3(2x + 6) = 2\log_3(1 - x)$
$\log_3 3 + \log_3(2x + 6) = \log_3(1 - x)^2$
$\log_3(3(2x + 6)) = \log_3(x^2 - 2x + 1)$
$\log_3(6x + 18) = \log_3(x^2 - 2x + 1)$

4. Equate the arguments and form a quadratic equation.

$6x + 18 = x^2 - 2x + 1$
$x^2 - 8x - 17 = 0$

5. Solve the quadratic equation and check for valid solutions.

$x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(1)(-17)}}{2(1)}$
$x = \frac{8 \pm \sqrt{64 + 68}}{2}$
$x = \frac{8 \pm \sqrt{132}}{2} = \frac{8 \pm 2\sqrt{33}}{2} = 4 \pm \sqrt{33}$

For the logarithms to be defined, we need $2x + 6 > 0 \implies x > -3$ and $1 - x > 0 \implies x < 1$.
Since $4 + \sqrt{33} \approx 9.74 > 1$, we reject $x = 4 + \sqrt{33}$.
Since $4 - \sqrt{33} \approx -1.74$, which satisfies $-3 < x < 1$, it is a valid solution.