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

We can use trigonometric identities to solve the equation. First, we know that $\cos^2 \theta + \sin^2 \theta = 1$, so we can write: \begin{align*} 2\sin^2 \theta &= 1 + \cos \theta \\ 2(1 - \cos^2 \theta) &= 1 + \cos \theta \\ 2 - 2\cos^2 \theta &= 1 + \cos \theta \\ 2\cos^2 \theta + \cos \theta - 1 &= 0 \\ (2\cos \theta - 1)(\cos \theta + 1) &= 0 \\ \end{align*} Therefore, either $2\cos \theta - 1 = 0$ or $\cos \theta + 1 = 0$. Solving for $\theta$ in each case, we get: \begin{align*} 2\cos \theta - 1 &= 0 \\ \cos \theta &= \frac{1}{2} \\ \theta &= 60^\circ \\ \\ \cos \theta + 1 &= 0 \\ \cos \theta &= -1 \\ \theta &= 180^\circ \\ \end{align*} However, we are given that $0^\circ \leq \theta \leq 90^\circ$, so the only valid solution is $\theta = 60^\circ$. Therefore, the answer is (C) 60°.