Suppose `x_1=tan^-1 2 >x_2>x_3>....` are the real numbers satisfying `sin(x_(n+1)-x_n)+2^-(n+1)*sin x_n*sinx_(n+1)` for all `n> 1` and `l=lim_(x->oo) x_n` , the value of `[4l]` is...where [t] denotes greatest integer function.

Answer

To solve the problem step by step, we will analyze the given equation and derive the limit \( l \) as \( n \) approaches infinity.

### Step 1: Understand the given equation
We are given that:
\[ \sin(x_{n+1} - x_n) + 2^{-(n+1)} \sin x_n \sin x_{n+1} = 0 \]
This implies:
\[ \sin(x_{n+1} - x_n) = -2^{-(n+1)} \sin x_n \sin x_{n+1} \]

### Step 2: Use the sine subtraction formula
Using the sine subtraction formula, we can rewrite the left-hand side:
\[ \sin(x_{n+1}) \cos(x_n) - \cos(x_{n+1}) \sin(x_n) = -2^{-(n+1)} \sin x_n \sin x_{n+1} \]

### Step 3: Rearranging the equation
Dividing both sides by \( \sin x_n \sin x_{n+1} \) (assuming they are non-zero), we get:
\[ \frac{\sin(x_{n+1}) \cos(x_n)}{\sin x_n \sin x_{n+1}} - \frac{\cos(x_{n+1}) \sin(x_n)}{\sin x_n \sin x_{n+1}} = -2^{-(n+1)} \]

### Step 4: Simplifying the equation
This simplifies to:
\[ \cot(x_n) - \cot(x_{n+1}) = 2^{-(n+1)} \]

### Step 5: Summing the differences
We can sum the differences from \( n = 1 \) to \( n = N \):
\[ \cot(x_1) - \cot(x_{N+1}) = \sum_{n=1}^{N} 2^{-(n+1)} \]
The right-hand side is a geometric series:
\[ \sum_{n=1}^{N} 2^{-(n+1)} = \frac{1/4(1 - (1/2)^N)}{1 - 1/2} = \frac{1}{2}(1 - (1/2)^N) \]

### Step 6: Finding the limit
As \( N \to \infty \):
\[ \cot(x_1) - \cot(x_{N+1}) = \frac{1}{2}(1 - 0) = \frac{1}{2} \]
Thus,
\[ \cot(x_{N+1}) \to \cot(x_1) - \frac{1}{2} \]

### Step 7: Substitute \( x_1 = \tan^{-1}(2) \)
Since \( x_1 = \tan^{-1}(2) \), we have:
\[ \cot(x_1) = \frac{1}{\tan(x_1)} = \frac{1}{2} \]
Therefore:
\[ \cot(x_{N+1}) \to \frac{1}{2} - \frac{1}{2} = 0 \]

### Step 8: Finding \( l \)
Thus, as \( n \to \infty \):
\[ \cot(x_{N+1}) \to 0 \implies x_{N+1} \to \frac{\pi}{2} \]
So, we have:
\[ l = \lim_{n \to \infty} x_n = \frac{\pi}{2} \]

### Step 9: Calculate \( 4l \)
Now, we calculate:
\[ 4l = 4 \cdot \frac{\pi}{2} = 2\pi \]

### Step 10: Apply the greatest integer function
The value of \( 2\pi \) is approximately \( 6.28 \), so:
\[ \lfloor 2\pi \rfloor = 6 \]

### Final Answer
Thus, the value of \( [4l] \) is:
\[ \boxed{6} \]