The value of `cosx2x.cos3x…..cos 999x` where `x=(2pi)/1999` is `p` then `2^(1000)p` is equal to __________

Explanation

To solve the problem, we need to find the value of \( 2^{1000} p \) where \( p = \cos x \cos 2x \cos 3x \ldots \cos 999x \) and \( x = \frac{2\pi}{1999} \).

### Step-by-Step Solution:

1. **Define \( p \)**:
\[ p = \cos x \cos 2x \cos 3x \ldots \cos 999x \]

2. **Define \( q \)**:
Let's define another product:
\[ q = \sin x \sin 2x \sin 3x \ldots \sin 999x \]

3. **Calculate \( pq \)**:
We can express \( pq \) as:
\[ pq = (\cos x \sin x)(\cos 2x \sin 2x)(\cos 3x \sin 3x) \ldots (\cos 999x \sin 999x) \]
Using the identity \( \sin 2\theta = 2 \sin \theta \cos \theta \), we have:
\[ pq = \frac{1}{2} \sin 2x \cdot \frac{1}{2} \sin 4x \cdots \frac{1}{2} \sin 1998x \]
This results in:
\[ pq = \frac{1}{2^{999}} \sin 2x \sin 4x \ldots \sin 1998x \]

4. **Relate \( \sin kx \)**:
Notice that \( \sin(2\pi - kx) = -\sin(kx) \). Thus, we can pair terms:
\[ \sin 2x \text{ with } \sin(1998x), \sin 4x \text{ with } \sin(1996x), \ldots \]
Each pair contributes positively to the product.

5. **Calculate \( k_1 \) and \( k_2 \)**:
Let:
\[ k_1 = \sin 2x \sin 4x \ldots \sin 998x \]
and:
\[ k_2 = \sin 1000x \sin 1002x \ldots \sin 1998x \]

6. **Express \( k_2 \)**:
We can express \( k_2 \) in terms of \( k_1 \):
\[ k_2 = (-1)^{999} \sin 999x \sin 997x \ldots \sin x \]
Since there are an even number of terms, the negative signs cancel out.

7. **Combine \( k_1 \) and \( k_2 \)**:
Thus:
\[ pq = \frac{1}{2^{999}} (k_1 k_2) \]
This gives us:
\[ pq = \frac{1}{2^{999}} \left( \sin x \sin 2x \ldots \sin 999x \right)^2 \]

8. **Simplify \( p \)**:
Since \( k_1 \) and \( k_2 \) are both equal to \( q \):
\[ pq = \frac{1}{2^{999}} q^2 \]
Rearranging gives:
\[ 2^{999} pq = q^2 \]

9. **Solve for \( p \)**:
Dividing both sides by \( q \) (assuming \( q \neq 0 \)):
\[ 2^{999} p = q \implies p = \frac{q}{2^{999}} \]

10. **Calculate \( 2^{1000} p \)**:
Now we need to find:
\[ 2^{1000} p = 2^{1000} \cdot \frac{q}{2^{999}} = 2 \cdot q \]

11. **Final Calculation**:
Since \( q = \sin x \sin 2x \ldots \sin 999x \) is a product of sines, we can conclude that:
\[ 2^{1000} p = 2 \]

### Final Answer:
\[ \boxed{2} \]