Solve the equation in the interval $\left(0^{\circ}, 360^{\circ}\right)$. Use an algebraic method. \[ 10 \sin ^{2} \theta-6 \sin \theta=3 \]

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, \(a = 10\), \(b = -6\) and \(c = -3\). 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 36.87 and 323.13. 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 \(\boxed{36.9^{\circ}, 323.1^{\circ}}\).