Using the definition of floor function yields that `floor[(3x+1)/2]` is the greatest integer `k <= x + 1` , hence, considering `x + 1 = k` , yields:

`x + 1 = k => x = k - 1`

Replacing `k - 1` for `x` in `floor` [`[(3x+1)/2]` yields:

`floor[(3x+1)/2]= floor[(3k - 3 + 1)/2] => floor[(3k - 2)/2] = k =>`

`k <= (3k - 2)/2 < k + 1 => {(k <= (3k - 2)/2),((3k - 2)/2 < k + 1):}`

`{(2k <= 3k - 2),(3k - 2 < 2k + 2):} => {(2 < 3k - 2k),(3k - 2k < 2 + 2):} => {(k>=2),(k<4):} => k in [2,oo) nn (-oo,4) nn Z`

`k in [2,4) nn Z => k in {2,3}`

You need to evaluate x, such that:

`{(x + 1 = k),(k=2):} => x = 2 - 1 = 1`

`{(x + 1 = k),(k=3):} => x = 3 - 1 = 2`

**Hence, evaluating the integer solutions to the equation that involves floor function, yields **`x = {1,2}.`