Solve the equation on the interval \( 0 \leq \theta<2 \pi \). \( 2 \cos (2 \theta)=6 \cos \theta-4 \) Select the correct choice below and fill in any answer boxes in your choice. A. \( \theta=\{ \} \) radians (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 needed.) B. There is no solution on this interval.

To solve the equation \( 2 \cos (2 \theta) = 6 \cos \theta - 4 \) on the interval \( 0 \leq \theta < 2\pi \), follow these steps: 1. **Use the double-angle identity for cosine:** \[ \cos(2\theta) = 2\cos^2\theta - 1 \] Substitute this into the original equation: \[ 2(2\cos^2\theta - 1) = 6\cos\theta - 4 \] Simplify: \[ 4\cos^2\theta - 2 = 6\cos\theta - 4 \] \[ 4\cos^2\theta - 6\cos\theta + 2 = 0 \] 2. **Solve the quadratic equation in terms of \( \cos\theta \):** \[ 4x^2 - 6x + 2 = 0 \quad \text{where} \quad x = \cos\theta \] Using the quadratic formula: \[ x = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 4 \cdot 2}}{2 \cdot 4} = \frac{6 \pm \sqrt{36 - 32}}{8} = \frac{6 \pm 2}{8} \] So, \( x = 1 \) or \( x = \frac{1}{2} \). 3. **Find \( \theta \) for each solution:** - For \( \cos\theta = 1 \): \[ \theta = 0 \quad \text{(since \( \cos(0) = 1 \))} \] - For \( \cos\theta = \frac{1}{2} \): \[ \theta = \frac{\pi}{3}, \frac{5\pi}{3} \] 4. **Combine the solutions:** \[ \theta = \left\{ 0, \frac{\pi}{3}, \frac{5\pi}{3} \right\} \] **Answer:** A. \( \theta=\left\{0,\ \dfrac{\pi}{3},\ \dfrac{5\pi}{3}\right\} \) radians