Solve the equation on the interval \( 0 \leq \theta<2 \pi \). \( 5 \sin ^{2} \theta-13 \sin \theta+8=0 \) 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 \{ \}. (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 \( 5 \sin^{2} \theta - 13 \sin \theta + 8 = 0 \), we can let \( x = \sin \theta \). This turns the equation into a quadratic form: \[ 5x^2 - 13x + 8 = 0. \] Applying the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): \[ x = \frac{13 \pm \sqrt{(-13)^2 - 4 \cdot 5 \cdot 8}}{2 \cdot 5} = \frac{13 \pm \sqrt{169 - 160}}{10} = \frac{13 \pm 3}{10}. \] This leads to two solutions for \( x \): 1. \( x = \frac{16}{10} = \frac{8}{5} \) (not valid since \( \sin \theta \) must be in the interval [-1, 1]), 2. \( x = \frac{10}{10} = 1 \). Now, for \( x = \sin \theta = 1 \), we have \( \theta = \frac{\pi}{2} \). Thus, the solution set in the interval \( 0 \leq \theta < 2\pi \) is: \(\{ \frac{\pi}{2} \}\).