Solve the equation. \[ 6-6 \sin \theta=4 \cos ^{2} \theta \] A. The solution set is \( \{ \) the solution in the interval \( 0 \leq \theta<2 \pi \) ? Select the correct choice and fill in any answer boxes in your choice below. (Simplify your answer. Type an exact answer, using \( \pi \) as needed. Type your answer in radians. Use integers or fractions for any to separate answers as needed.) B. There is no solution.

Solve the equation \( 6-6 \sin \theta=4 \cos ^{2} \theta \). Solve the equation by following steps: - step0: Solve for \(\theta\): \(6-6\sin\left(\theta \right)=4\cos^{2}\left(\theta \right)\) - step1: Rewrite the expression: \(6-6\sin\left(\theta \right)=4-4\sin^{2}\left(\theta \right)\) - step2: Move the expression to the left side: \(6-6\sin\left(\theta \right)-\left(4-4\sin^{2}\left(\theta \right)\right)=0\) - step3: Calculate: \(2-6\sin\left(\theta \right)+4\sin^{2}\left(\theta \right)=0\) - step4: Factor the expression: \(2\left(1-\sin\left(\theta \right)\right)\left(1-2\sin\left(\theta \right)\right)=0\) - step5: Calculate: \(\left(1-\sin\left(\theta \right)\right)\left(1-2\sin\left(\theta \right)\right)=0\) - step6: Separate into possible cases: \(\begin{align}&1-\sin\left(\theta \right)=0\\&1-2\sin\left(\theta \right)=0\end{align}\) - step7: Solve the equation: \(\begin{align}&\theta =\frac{\pi }{2}+2k\pi ,k \in \mathbb{Z}\\&\theta =\left\{ \begin{array}{l}\frac{\pi }{6}+2k\pi \\\frac{5\pi }{6}+2k\pi \end{array}\right.,k \in \mathbb{Z}\end{align}\) - step8: Find the union: \(\theta =\left\{ \begin{array}{l}\frac{\pi }{6}+2k\pi \\\frac{\pi }{2}+2k\pi \\\frac{5\pi }{6}+2k\pi \end{array}\right.,k \in \mathbb{Z}\) The solution to the equation \(6-6 \sin \theta=4 \cos ^{2} \theta\) is given by: \[ \theta = \left\{ \frac{\pi}{6}+2k\pi, \frac{\pi}{2}+2k\pi, \frac{5\pi}{6}+2k\pi \right\}, \text{where } k \in \mathbb{Z} \] Therefore, the solution set is \( \{ \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6} \} \) in the interval \( 0 \leq \theta < 2\pi \).