If \(2\sin^{2}\theta = 1 + \cos \theta, 0° \leq \theta \leq 90°\), find \(\theta\).

Explanation

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

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

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

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

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

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

\(\theta = \cos^{-1} \frac{1}{2} = 60°\)