Start with the equation:

\(\frac{\tan \theta + 2 \sin \theta}{\tan \theta - 2 \sin \theta} = 3\)

Cross-multiply to get:

\(\tan \theta + 2 \sin \theta = 3(\tan \theta - 2 \sin \theta)\)

Expand the right side:

\(\tan \theta + 2 \sin \theta = 3 \tan \theta - 6 \sin \theta\)

Rearrange terms:

\(2 \tan \theta - 8 \sin \theta = 0\)

Factor out \(2 \sin \theta\):

\(2 \sin \theta (\tan \theta - 4 \cos \theta) = 0\)

Since \(\sin \theta = 0\) is not valid in the given range, solve:

\(\tan \theta = 4 \cos \theta\)

Using \(\tan \theta = \frac{\sin \theta}{\cos \theta}\), we have:

\(\frac{\sin \theta}{\cos \theta} = 4 \cos \theta\)

\(\sin \theta = 4 \cos^2 \theta\)

Using \(\cos^2 \theta = 1 - \sin^2 \theta\), substitute:

\(\sin \theta = 4(1 - \sin^2 \theta)\)

\(\sin \theta = 4 - 4 \sin^2 \theta\)

Rearrange to form a quadratic equation:

\(4 \sin^2 \theta + \sin \theta - 4 = 0\)

Using the quadratic formula, solve for \(\sin \theta\):

\(\sin \theta = \frac{-1 \pm \sqrt{1 + 64}}{8}\)

\(\sin \theta = \frac{-1 \pm 9}{8}\)

\(\sin \theta = 1\) or \(\sin \theta = -\frac{5}{4}\)

Only \(\sin \theta = 1\) is valid, leading to \(\theta = 90^\circ\), but this is not in the solution range.

Thus, solve \(\cos \theta = \frac{1}{4}\):

\(\theta = 75.5^\circ\) only.