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 given problem, we need to analyze the equations step by step.

### Given Equations:
1. \( x + \sin y = 2009 \) (Equation 1)
2. \( x + 2009 \cos y = 2008 \) (Equation 2)

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

### Step 2: Rearranging Equation 2
From Equation 2, we can also express \( x \):
\[ x = 2008 - 2009 \cos y \]

### Step 3: Setting the two expressions for \( x \) equal
Since both expressions represent \( x \), we can set them equal to each other:
\[ 2009 - \sin y = 2008 - 2009 \cos y \]

### Step 4: Simplifying the equation
Rearranging the equation gives:
\[ 2009 - 2008 = -\sin y + 2009 \cos y \]
\[ 1 = -\sin y + 2009 \cos y \]

### Step 5: Rearranging to isolate \(\sin y\)
Rearranging gives:
\[ \sin y = 2009 \cos y - 1 \]

### Step 6: Analyzing the equation
For the equation \(\sin y = 2009 \cos y - 1\) to hold, we need to consider the range of \(\sin y\) and \(\cos y\). Since \(y\) is in the interval \([0, \frac{\pi}{2}]\):
- \(\sin y\) ranges from 0 to 1.
- \(\cos y\) ranges from 1 to 0.

### Step 7: Finding conditions for \(\cos y\)
The equation \(\sin y = 2009 \cos y - 1\) implies:
\[ 2009 \cos y - 1 \geq 0 \implies 2009 \cos y \geq 1 \implies \cos y \geq \frac{1}{2009} \]

### Step 8: Finding when \(\sin y\) equals 1
The maximum value of \(\sin y\) is 1, which occurs when \(y = \frac{\pi}{2}\). At this point, \(\cos y = 0\):
\[ \sin\left(\frac{\pi}{2}\right) = 1 \]
Substituting \(y = \frac{\pi}{2}\) into the rearranged equation:
\[ 1 = 2009 \cdot 0 - 1 \quad \text{(not valid)} \]

### Step 9: Setting \(\cos y = 0\)
The only valid case occurs when \(\cos y = 0\) (which happens at \(y = \frac{\pi}{2}\)):
\[ \sin\left(\frac{\pi}{2}\right) = 1 \]
Thus, substituting \(y = \frac{\pi}{2}\) into Equation 1:
\[ x + 1 = 2009 \implies x = 2008 \]

### Step 10: Finding \(x + y\)
Now, we find \(x + y\):
\[ x + y = 2008 + \frac{\pi}{2} \]

### Step 11: Applying the greatest integer function
To find \([x + y]\), we need to evaluate:
\[ [x + y] = \left[2008 + \frac{\pi}{2}\right] \]
Since \(\frac{\pi}{2} \approx 1.57\), we have:
\[ 2008 + 1.57 \approx 2009.57 \]
Thus,
\[ [x + y] = 2009 \]

### Final Answer:
\[ [x + y] = 2009 \]