$\sin ^{2}(x)+\sin (x)-1=0$

Step 1 ：Given the equation \(\sin ^{2}(x)+\sin (x)-1=0\), this is a quadratic equation in terms of \(\sin(x)\).

Step 2 ：We can solve it by using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where a, b, and c are the coefficients of the quadratic equation. In this case, a = 1, b = 1, and c = -1.

Step 3 ：By substituting the values of a, b, and c into the quadratic formula, we get two solutions for \(\sin(x)\): -1.618033988749895 and 0.6180339887498949.

Step 4 ：However, the range of \(\sin(x)\) is from -1 to 1. Therefore, -1.618033988749895 is not a valid solution. We only consider 0.6180339887498949 as the solution for \(\sin(x)\).

Step 5 ：Now, we need to find the value of x that makes \(\sin(x) = 0.6180339887498949\). By using the inverse sine function, we find that x = 38.17270762701225 degrees.

Step 6 ：Final Answer: The solution to the equation \(\sin ^{2}(x)+\sin (x)-1=0\) is \(\boxed{38.17}\) degrees.