This question: 1 point(s) possible Submit quiz Solve the equation on the interval $0 \leq \theta<2 \pi$. \[ \tan (2 \theta)+2 \cos \theta=0 \] Select the correct choice below and fill in any answer boxes in your choice. A. The solution set is (Use a comma to separate answers as needed. Type an exact answer, using $\pi$ as needed. Use integers or fractions for any numbers in the expression.) B. There is no solution.

Step 1 ：\(\tan(2\theta) = \frac{2\sin\theta\cos\theta}{\cos^2\theta-\sin^2\theta} = \frac{\sin(2\theta)}{\cos(2\theta)}\)

Step 2 ：\(\sin(2\theta) = 2\sin\theta\cos\theta\) and \(\cos(2\theta) = \cos^2\theta-\sin^2\theta = 2\cos^2\theta - 1\)

Step 3 ：\(\tan(2\theta) = \frac{2\sqrt{1-\cos^2\theta}\cos\theta}{2\cos^2\theta - 1}\)

Step 4 ：Substitute \(\tan(2\theta)\) into the original equation: \(\frac{2\sqrt{1-\cos^2\theta}\cos\theta}{2\cos^2\theta - 1} + 2\cos\theta = 0\)

Step 5 ：Multiply through by \(2\cos^2\theta - 1\): \(2\sqrt{1-\cos^2\theta}\cos\theta + 2\cos\theta(2\cos^2\theta - 1) = 0\)

Step 6 ：Simplify: \(2\sqrt{1-\cos^2\theta}\cos\theta + 4\cos^3\theta - 2\cos\theta = 0\)

Step 7 ：Rearrange: \(4\cos^3\theta - 2\cos\theta + 2\sqrt{1-\cos^2\theta}\cos\theta = 0\)

Step 8 ：Factor out \(2\cos\theta\): \(2\cos\theta(2\cos^2\theta - 1 + \sqrt{1-\cos^2\theta}) = 0\)

Step 9 ：Set each factor equal to zero: \(2\cos\theta = 0\) and \(2\cos^2\theta - 1 + \sqrt{1-\cos^2\theta} = 0\)

Step 10 ：Solve for \(\cos\theta\) in the first equation: \(\cos\theta = 0\)

Step 11 ：Solve for \(\theta\) in the first equation: \(\theta = \frac{\pi}{2}\) or \(\theta = \frac{3\pi}{2}\)

Step 12 ：The second equation has no real solutions because the discriminant is negative

Step 13 ：\(\boxed{\theta = \frac{\pi}{2}, \frac{3\pi}{2}}\)