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??????

Den i hv 2 think abt sum other way! wat's the ans??

(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]

i thnk answer wuld be x → (-∞, -2) U [-1, 2]

CASE : I
4 - x² ≤ 0 → x E ( -∞ , -2] U [2 ,∞)
and [x] + 2 < 0 → x E ( -∞, -2)
hence in all
x E ( - ∞ , -2)

CASE II
4 - x² ≥ 0 → x E [ -2 , 2 ]
and
[x] + 2 > 0 → x E [-1 , ∞ )
hence in all
x E [-1 , 2]

combining CASE I and CASE II
we get
x E ( -∞ , -2 ) U [ -1 , 2 ]