For the smallest positive values of `xa n dy ,` the equation `2(s inx+s in y)-2cos(x-y)=3` has a solution, then which of the following is/are true? `sin(x+y)/2=1` (b) `cos((x-y)/2)=1/2` number of ordered pairs `(x , y)` is 2 number of ordered pairs `(x , y)i s3`

Options

A. `"sin" (x+y)/2=1`
B. `cos((x-y)/2)=1/2`
C. number of ordered pairs (x, y) is 2
D. number of ordered pairs (x, y) is 3

Explanation

To solve the equation \( 2(\sin x + \sin y) - 2\cos(x - y) = 3 \) for the smallest positive values of \( x \) and \( y \), we will follow these steps:

### Step 1: Rewrite the equation
We start with the given equation:
\[ 2(\sin x + \sin y) - 2\cos(x - y) = 3 \]
Dividing the entire equation by 2 gives:
\[ \sin x + \sin y - \cos(x - y) = \frac{3}{2} \]

### Step 2: Use trigonometric identities
Using the identity for the sum of sines:
\[ \sin x + \sin y = 2 \sin\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right) \]
We can rewrite the equation as:
\[ 2 \sin\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right) - \cos(x - y) = \frac{3}{2} \]

### Step 3: Rewrite \(\cos(x - y)\)
Using the identity for cosine:
\[ \cos(x - y) = 2\cos^2\left(\frac{x-y}{2}\right) - 1 \]
Substituting this into the equation gives:
\[ 2 \sin\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right) - (2\cos^2\left(\frac{x-y}{2}\right) - 1) = \frac{3}{2} \]

### Step 4: Rearranging the equation
Rearranging the equation leads to:
\[ 2 \sin\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right) + 1 - 2\cos^2\left(\frac{x-y}{2}\right) = \frac{3}{2} \]
This simplifies to:
\[ 2 \sin\left(\frac{x+y}{2}\right) \cos\left(\frac{x-y}{2}\right) - 2\cos^2\left(\frac{x-y}{2}\right) = \frac{1}{2} \]

### Step 5: Setting up conditions
For the equation to have solutions, we analyze the conditions:
1. **For \( \sin\left(\frac{x+y}{2}\right) = 1 \)**:
This implies:
\[ \frac{x+y}{2} = \frac{\pi}{2} \implies x + y = \pi \]

2. **For \( \cos\left(\frac{x-y}{2}\right) = \frac{1}{2} \)**:
This implies:
\[ \frac{x-y}{2} = \frac{\pi}{3} \text{ or } \frac{5\pi}{3} \implies x - y = \frac{2\pi}{3} \text{ or } -\frac{2\pi}{3} \]

### Step 6: Solving the system of equations
Now we have two equations:
1. \( x + y = \pi \)
2. \( x - y = \frac{2\pi}{3} \)

Adding these equations:
\[ 2x = \pi + \frac{2\pi}{3} \implies 2x = \frac{3\pi + 2\pi}{3} = \frac{5\pi}{3} \implies x = \frac{5\pi}{6} \]
Substituting \( x \) back into \( x + y = \pi \):
\[ \frac{5\pi}{6} + y = \pi \implies y = \pi - \frac{5\pi}{6} = \frac{\pi}{6} \]

### Step 7: Finding ordered pairs
We can also consider the negative solution for \( x - y \):
1. \( x - y = -\frac{2\pi}{3} \)

Adding this to \( x + y = \pi \):
\[ 2x = \pi - \frac{2\pi}{3} \implies 2x = \frac{3\pi - 2\pi}{3} = \frac{\pi}{3} \implies x = \frac{\pi}{6} \]
Substituting back gives:
\[ \frac{\pi}{6} + y = \pi \implies y = \frac{5\pi}{6} \]

### Conclusion
Thus, the ordered pairs \( (x, y) \) are:
1. \( \left(\frac{5\pi}{6}, \frac{\pi}{6}\right) \)
2. \( \left(\frac{\pi}{6}, \frac{5\pi}{6}\right) \)

### Final Answers
- \( \sin\left(\frac{x+y}{2}\right) = 1 \) is **True**.
- \( \cos\left(\frac{x-y}{2}\right) = \frac{1}{2} \) is **True**.
- The number of ordered pairs \( (x, y) \) is **2**.