First, separate the variables:

\(\int \tan \theta \, \mathrm{d}\theta = \int \frac{x}{x^2 + 3} \, \mathrm{d}x.\)

Integrate both sides:

\(-\ln(\cos \theta) = \frac{1}{2} \ln(x^2 + 3) + C.\)

Use the initial condition \(x = 1\) when \(\theta = 0\) to find \(C\):

\(-\ln(1) = \frac{1}{2} \ln(1^2 + 3) + C\)

\(0 = \frac{1}{2} \ln(4) + C\)

\(C = -\ln(2)\)

Substitute back to get:

\(-\ln(\cos \theta) = \frac{1}{2} \ln(x^2 + 3) - \ln(2)\)

Rearrange to solve for \(x^2\):

\(\ln(\cos \theta) + \ln(2) = \frac{1}{2} \ln(x^2 + 3)\)

\(\ln(2 \cos \theta) = \frac{1}{2} \ln(x^2 + 3)\)

\(2 \ln(2 \cos \theta) = \ln(x^2 + 3)\)

\(x^2 + 3 = 4 \cos^2 \theta\)

\(x^2 = \frac{4}{\cos^2 \theta} - 3\)