To solve this inequality, we can first factor out sin x:

sin x(2sin x + 1) ≥ 0

Now we need to find the values of x that satisfy this inequality. There are three cases to consider:

Case 1: sin x = 0
If sin x = 0, then the inequality simplifies to:
0(2(0) + 1) ≥ 0
0 ≥ 0

Since 0 is greater than or equal to 0, this case is true for all values of x where sin x = 0.

Case 2: 2sin x + 1 = 0
If 2sin x + 1 = 0, then sin x = -1/2. The solutions for this case will be the x values where sin x = -1/2.

Case 3: sin x(2sin x + 1) > 0
If sin x and 2sin x + 1 have the same sign (both positive or both negative), then their product is positive.

For sin x > 0 and 2sin x + 1 > 0:
sin x > 0 and sin x > -1/2
This is true for x in the first and second quadrant.

For sin x < 0 and 2sin x + 1 < 0:
sin x < 0 and sin x < -1/2
This is true for x in the third and fourth quadrant.

Therefore, the solution to the inequality 2sin^2 x + sin x ≥ 0 is:
x ∈ [0, π] ∪ [2π, 3π] ∪ ...

or in interval notation:
[0, π] ∪ [2π, 3π] ∪ ...