\(2\sin^{2} \theta = 1 + \cos \theta\)

\(2 ( 1 - \cos^{2} \theta) = 1 + \cos \theta\)

\(2 - 2\cos^{2} \theta = 1 + \cos \theta\)

\(0 = 1 - 2 + \cos \theta + 2\cos^{2} \theta\)

\(2\cos^{2} \theta + \cos \theta - 1 = 0\)

Factorizing, we have

\((\cos \theta + 1)(2\cos \theta - 1) = 0\)

Note: In the range, \(0° \leq \theta \leq 90°\), all trig functions are positive, so we consider

\(2\cos \theta = 1 \implies \cos \theta = \frac{1}{2}\)

\(\theta = 60°\).