Solve the equation on the interval \( 0 \leq \theta<2 \pi \). \[ \cos (2 \theta)+6 \sin ^{2} \theta=4 \] What 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. A. \( \theta= \) \( \square \) ) (Simplify your answer. Type an exact answer, using \( \pi \) as needed. Type your answer in radians. Use integers or fractions for any numbers in the expression. Use a comma to separate answers as noeded.) B. There is no solution.

To solve the equation \(\cos (2 \theta) + 6 \sin^2 \theta = 4\), we start by using the double angle identity, \(\cos(2\theta) = 1 - 2\sin^2(\theta)\). Substituting that into the equation gives: \[ 1 - 2\sin^2 \theta + 6\sin^2 \theta = 4 \] This simplifies to: \[ 1 + 4\sin^2 \theta = 4 \] Subtracting 1 from both sides results in: \[ 4\sin^2 \theta = 3 \] Dividing both sides by 4 yields: \[ \sin^2 \theta = \frac{3}{4} \] Taking the square root of both sides, we find: \[ \sin \theta = \pm \frac{\sqrt{3}}{2} \] Now we need to find the angles \(\theta\) where \(\sin \theta = \frac{\sqrt{3}}{2}\) and \(\sin \theta = -\frac{\sqrt{3}}{2}\): 1. For \(\sin \theta = \frac{\sqrt{3}}{2}\): - \(\theta = \frac{\pi}{3}, \frac{2\pi}{3}\) 2. For \(\sin \theta = -\frac{\sqrt{3}}{2}\): - \(\theta = \frac{4\pi}{3}, \frac{5\pi}{3}\) Putting it all together, the solutions in the interval \(0 \leq \theta < 2\pi\) are: \(\theta = \frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}\) So the correct choice is: A. \( \theta= \frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3} \)