To solve the differential equation \(\frac{dx}{dθ} = (x + 2) \sin^2 2θ\), we first separate variables:

\(\frac{1}{x + 2} dx = \sin^2 2θ \, dθ\).

Integrating both sides, we have:

\(\int \frac{1}{x + 2} \, dx = \int \sin^2 2θ \, dθ\).

The left side integrates to \(\ln(x + 2)\).

For the right side, use the identity \(\sin^2 2θ = \frac{1 - \cos 4θ}{2}\):

\(\int \sin^2 2θ \, dθ = \int \frac{1 - \cos 4θ}{2} \, dθ = \frac{1}{2} \int (1 - \cos 4θ) \, dθ\).

This integrates to \(\frac{1}{2} \left( θ - \frac{1}{4} \sin 4θ \right)\).

Thus, we have:

\(\ln(x + 2) = \frac{1}{2} θ - \frac{1}{8} \sin 4θ + C\).

Using the initial condition \(x = 0\) when \(θ = 0\), we find:

\(\ln(2) = C\).

So the solution is:

\(\ln(x + 2) = \frac{1}{2} θ - \frac{1}{8} \sin 4θ + \ln 2\).

To find \(x\) when \(θ = \frac{1}{4}π\):

\(\ln(x + 2) = \frac{1}{2} \left( \frac{1}{4}π \right) - \frac{1}{8} \sin \left( π \right) + \ln 2\).

\(\ln(x + 2) = \frac{π}{8} + \ln 2\).

Solving for \(x\):

\(x + 2 = e^{\frac{π}{8} + \ln 2} = 2e^{\frac{π}{8}}\).

\(x = 2e^{\frac{π}{8}} - 2\).

Calculating this gives \(x \approx 0.962\) to 3 significant figures.