Domain Answer not matching

1st of all: 4-x²â‰¥0
i.e,(x+2)(x-2)≤0
so,the value of x varies as: -2≤x≤2

Now,[x]+2>0 {N.B: Not ≥0}
i.e, [x]>-2
i.e, n-1>-2
i.e, n>-1
so,the value of [x] will vary 4m: (-1,0]U[0,1)U[1,2)...........

so the ans is: xε(-1,0)Ï
[0,1)Ï
[1,2)

dis is juss a GUESS!!!
ans match korlo??????

(EDITED on Manmay's request)
Go by the restrictions imposed by different functions we have...oi shob restriction er intersection nile tumi domain paabe.

R1 : Denominator cannot be zero

[x] + 2 ≠0
=> [x] ≠-2

Now [x] = -2 for -2 ≤ x < -1
So [x] ≠-2 for x < -2 U x ≥ -1

Let me explain. For [x] = 2, we are given that x ≥ -2 AND x < -1. Let (-2, ∞) = A and (-∞, -1] = B. So our expression becomes (A ∩ B) where ∩ is intersection/AND. Now when we have to find for [x] ≠2, we complement that set.
By DeMorgan's law, (A ∩ B)' = A' U B' which is the set I found.

So we have x → (-∞, -2) U [-1, ∞)

R2 : Square root term must be non negative

4 - x²[x] + 2 ≥ 0

We have two subcases here.
CASE I : N^r ≥ 0 AND D^r > 0
So x² ≤ 4 AND [x] > -2

Here you must know about the Greatest Integer inequalities. I posted them here : http://www.targetiit.com/iit-jee-forum/posts/greatest-integer-function-exhaustive-collection-pl-13636.html

[x] > I implies x ≥ I + 1
So [x] > -2 implies x ≥ -1

So |x| ≤ 2 AND x ≥ -1
=> -2 ≤ x ≤ 2 AND x ≥ -1
Therefore x → [-1, 2]

or we have CASE II : N^r ≤ 0 AND D^r < 0
|x| ≥ 2 AND [x] < -2
=> x ≤ -2 U x ≥ 2 AND x < -2 (as [x] < I implies x < I)
=> x → (-∞, -2)

So our final restriction(2nd) is x→ (-∞, -2) U [-1, 2] (UNION of both cases)

Now our domain of function is obtained by :
D_f = R1 ∩ R2
= x → (-∞, -2) U [-1, ∞) ∩ (-∞, -2) U [-1, 2]
So our final domain is x → (-∞, -2) U [-1, 2]