(C) Given $\sin^2 \theta = \frac{1}{4}.$We know that $\sin^2 \theta = \sin^2 \alpha \implies \theta = n\pi \pm \alpha.$Since $\sin^2 \theta = \frac{1}

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(C) Given $\sin^2 \theta = \frac{1}{4}.$We know that $\sin^2 \theta = \sin^2 \alpha \implies \theta = n\pi \pm \alpha.$Since $\sin^2 \theta = \frac{1}{4} = \left(\frac{1}{2}\right)^2 = \sin^2 \left(\frac{\pi}{6}\right),$we have $\alpha = \frac{\pi}{6}.$Therefore,the general solution is $\theta = n\pi \pm \frac{\pi}{6},$ where $n \in \mathbb{Z}.$