What is the solution to the trigonometric inequality 2sin(x) + 3sin^2 (x) over the interval 0 <= x <= 2pi radians?

e

step-by-step explanation:

an equation by definition has to have two or more variables (x and 2 in this case)

two? xd

step-by-step explanation:

answer: 5

step-by-step explanation:

since, the range is the difference between the lowest and highest values of the given sample space.

and, here the sample space = { 3, 1, 0, 4, 1, 4, 2, 3, 0, 5, 0, 1, 1, 2, 0, 4, 1,4, 4 ,2 ,3 ,2 ,0 ,1 ,4 }

here, the highest value of the given sample = 5

and, the lowest value = 0

therefore, the range of the sample space = 5 - 0 = 5