To solve this quadratic equation for sin(x), we can use the quadratic formula:

sin(x) = [-b ± √(b^2 - 4ac)] / 2a

where a = 4, b = 11, and c = -3 in the equation ax^2 + bx + c = 0.

Plugging these values into the formula, we get:

sin(x) = [-11 ± √(11^2 - 44(-3))] / 2*4
sin(x) = [-11 ± √(121 + 48)] / 8
sin(x) = [-11 ± √169] / 8
sin(x) = [-11 ± 13] / 8

Now, we can find the two possible solutions for sin(x):

sin(x) = (-11 + 13) / 8 = 2 / 8 = 0.25
Taking the arcsin of 0.25 gives us x = arcsin(0.25) ≈ 14.48 degrees

sin(x) = (-11 - 13) / 8 = -24 / 8 = -3
Since the sine function has a range of [-1, 1], this solution is outside the valid range for sin(x) and is therefore extraneous.

Therefore, the solution to the equation 4sin^2x + 11 sinx - 3 = 0 is sin(x) = 0.25, or x ≈ 14.48 degrees.