Points: 0 of 1 Solve the equation on the interval \( 0 \leq \theta<2 \pi \). \[ 3 \sin ^{2} \theta+10 \sin \theta+7=0 \] 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. The solution set is \( \square \) \( \} \). (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 needed.) B. There is no solution.

To solve the equation \(3 \sin^2 \theta + 10 \sin \theta + 7 = 0\), we can start by letting \(x = \sin \theta\). This transforms our equation into a quadratic form: \[ 3x^2 + 10x + 7 = 0. \] Now, we can apply the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 3\), \(b = 10\), and \(c = 7\). Calculating the discriminant: \[ b^2 - 4ac = 10^2 - 4 \cdot 3 \cdot 7 = 100 - 84 = 16. \] Now, plugging in the values into the quadratic formula: \[ x = \frac{-10 \pm \sqrt{16}}{2 \cdot 3} = \frac{-10 \pm 4}{6}. \] This gives us two possible solutions for \(x\): 1. \(x = \frac{-10 + 4}{6} = \frac{-6}{6} = -1\) 2. \(x = \frac{-10 - 4}{6} = \frac{-14}{6} = -\frac{7}{3}\) Since \(-\frac{7}{3}\) is not a valid sine value (as \( \sin \theta \) must be between \(-1\) and \(1\)), we only have one possible solution: 1. \( \sin \theta = -1 \) The sine equals \(-1\) at \[ \theta = \frac{3\pi}{2}. \] Thus, the solution in the interval \(0 \leq \theta < 2\pi\) is A. The solution set is \(\{\frac{3\pi}{2}\}\).