The first term \(a = \cos \theta\) and the second term \(a + d = \cos \theta + \sin^2 \theta\) gives the common difference \(d = \sin^2 \theta\).

The sum of the first 13 terms of an arithmetic progression is given by:

\(S_{13} = \frac{13}{2} \left[ 2a + 12d \right] = 52\)

Substitute \(a = \cos \theta\) and \(d = \sin^2 \theta\):

\(\frac{13}{2} \left[ 2\cos \theta + 12\sin^2 \theta \right] = 52\)

Simplify and solve for \(\theta\):

\(2\cos \theta + 12(1 - \cos^2 \theta) = 8\)

\(2\cos \theta + 12 - 12\cos^2 \theta = 8\)

\(12\cos^2 \theta - 2\cos \theta - 4 = 0\)

Divide by 2:

\(6\cos^2 \theta - \cos \theta - 2 = 0\)

Solving the quadratic equation for \(\cos \theta\):

\(\cos \theta = \frac{2}{3} \text{ or } \cos \theta = -\frac{1}{2}\)

For \(\cos \theta = \frac{2}{3}\), \(\theta \approx 0.841\).

For \(\cos \theta = -\frac{1}{2}\), \(\theta \approx 2.09\).