Given x + sin y = 2009 and x + 2009 cos y = 2008, where `y in[0,(pi)/(2)]`, then [x+y] equals, where [.] represents the greatest integer function,

Options

A. 2008
B. 2009
C. 2100
D. 2010

Explanation

To solve the problem, we have the following equations:

1. \( x + \sin y = 2009 \) (Equation 1)
2. \( x + 2009 \cos y = 2008 \) (Equation 2)

We need to find \( [x + y] \), where \( [.] \) represents the greatest integer function.

### Step 1: Subtract Equation 2 from Equation 1

From Equation 1, we can express \( x \) in terms of \( \sin y \):
\[ x = 2009 - \sin y \]

From Equation 2, we can express \( x \) in terms of \( \cos y \):
\[ x = 2008 - 2009 \cos y \]

Setting these two expressions for \( x \) equal to each other:
\[ 2009 - \sin y = 2008 - 2009 \cos y \]

### Step 2: Rearranging the equation

Rearranging gives:
\[ \sin y - 2009 \cos y = 1 \]

### Step 3: Analyze the equation

We know that \( \sin y \) can take values from \( 0 \) to \( 1 \) for \( y \in [0, \frac{\pi}{2}] \). Therefore, we can express \( \sin y \) in terms of \( \cos y \):
\[ \sin y = 1 + 2009 \cos y \]

### Step 4: Determine the implications

Since \( \sin y \) must be less than or equal to \( 1 \):
\[ 1 + 2009 \cos y \leq 1 \]
This implies:
\[ 2009 \cos y \leq 0 \]
Given that \( \cos y \) is non-negative in the interval \( [0, \frac{\pi}{2}] \), the only solution is:
\[ \cos y = 0 \]

### Step 5: Solve for \( y \)

The only value of \( y \) in the interval \( [0, \frac{\pi}{2}] \) for which \( \cos y = 0 \) is:
\[ y = \frac{\pi}{2} \]

### Step 6: Find \( x \)

Substituting \( y = \frac{\pi}{2} \) back into Equation 1:
\[ x + \sin\left(\frac{\pi}{2}\right) = 2009 \]
\[ x + 1 = 2009 \]
\[ x = 2008 \]

### Step 7: Calculate \( x + y \)

Now, we can find \( x + y \):
\[ x + y = 2008 + \frac{\pi}{2} \]

### Step 8: Find the greatest integer function

To find \( [x + y] \):
\[ [x + y] = [2008 + \frac{\pi}{2}] \]
Since \( \frac{\pi}{2} \approx 1.57 \):
\[ 2008 + \frac{\pi}{2} \approx 2009.57 \]
Thus,
\[ [x + y] = 2009 \]

### Final Answer:
\[ \boxed{2009} \]