Solve the equation in the interval $\left[0^{\circ}, 360^{\circ}\right)$. Use an algebraic method. $11 \sin ^{2} \theta-6 \sin \theta=4$ Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The solution set is (Simplify your answer. Round to the nearest tenth as needed. Use a comma to separate answers as needed. Do not include the degree symbol in your answer) B. The solution is the empty set

Step 1 ：This is a quadratic equation in terms of \(\sin \theta\). We can solve it by using the quadratic formula. The quadratic formula is given by: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a\), \(b\) and \(c\) are the coefficients of the quadratic equation \(ax^2 + bx + c = 0\). In this case, \(a = 11\), \(b = -6\) and \(c = -4\). We can substitute these values into the quadratic formula to find the solutions for \(\sin \theta\).

Step 2 ：After finding the solutions for \(\sin \theta\), we can use the inverse sine function to find the values of \(\theta\) in the interval \([0^{\circ}, 360^{\circ})\).

Step 3 ：The solutions for \(\theta\) are approximately 69.16 and 337.10. These are the angles in degrees that satisfy the given equation in the interval \([0^{\circ}, 360^{\circ})\). We should round these values to the nearest tenth as instructed in the question.

Step 4 ：Final Answer: The solution set is \(\boxed{69.2^{\circ}, 337.1^{\circ}}\).