Separate variables:

\(\int \frac{1}{x+2} \, dx = \int \cot \frac{1}{2} \theta \, d\theta\)

Integrate both sides:

\(\ln(x+2) = 2 \ln \sin \frac{1}{2} \theta + C\)

Use the initial condition \(x = 1\) when \(\theta = \frac{1}{3} \pi\):

\(\ln(1+2) = 2 \ln \sin \frac{1}{6} \pi + C\)

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

\(C = \ln 3 - 2 \ln \frac{1}{2} = \ln 12\)

Substitute back:

\(\ln(x+2) = 2 \ln \sin \frac{1}{2} \theta + \ln 12\)

\(x+2 = 12 \sin^2 \frac{1}{2} \theta\)

Use the double angle identity \(\sin^2 \frac{1}{2} \theta = \frac{1 - \cos \theta}{2}\):

\(x+2 = 12 \cdot \frac{1 - \cos \theta}{2}\)

\(x+2 = 6(1 - \cos \theta)\)

\(x = 6 - 6 \cos \theta - 2\)

\(x = 4 - 6 \cos \theta\)