Consider the system of equations
`sin x cos 2y=(a^(2)-1)^(2)+1, cos x sin 2y = a+1`
The number of values of `y in [0, 2pi]`, when the system has solution for permissible values of a, are

Options

A. 1
B. 2
C. 3
D. 4

Explanation

To solve the given system of equations:

1. **Equations Given:**
\[ \sin x \cos 2y = (a^2 - 1)^2 + 1 \quad (1) \]
\[ \cos x \sin 2y = a + 1 \quad (2) \]

2. **Analyzing the Left-Hand Side (LHS):**
The LHS of both equations, \(\sin x \cos 2y\) and \(\cos x \sin 2y\), must be bounded between -1 and 1, since both sine and cosine functions have a range of [-1, 1]. Therefore, we have:
\[ -1 \leq (a^2 - 1)^2 + 1 \leq 1 \quad (3) \]
\[ -1 \leq a + 1 \leq 1 \quad (4) \]

3. **Solving Inequalities:**
From inequality (3):
\[ (a^2 - 1)^2 + 1 \leq 1 \implies (a^2 - 1)^2 \leq 0 \]
This implies:
\[ (a^2 - 1)^2 = 0 \implies a^2 - 1 = 0 \implies a^2 = 1 \implies a = 1 \text{ or } a = -1 \]

From inequality (4):
\[ -1 \leq a + 1 \leq 1 \implies -2 \leq a \leq 0 \]

4. **Finding Permissible Values of \(a\):**
The permissible values of \(a\) from both inequalities are:
- From (3): \(a = 1\) or \(a = -1\)
- From (4): \(-2 \leq a \leq 0\)

The only value that satisfies both conditions is:
\[ a = -1 \]

5. **Substituting \(a = -1\) into the Equations:**
Substitute \(a = -1\) into equations (1) and (2):
\[ \sin x \cos 2y = (1 - 1)^2 + 1 = 1 \quad (5) \]
\[ \cos x \sin 2y = -1 + 1 = 0 \quad (6) \]

6. **Analyzing Equation (5):**
From equation (5):
\[ \sin x \cos 2y = 1 \]
This means:
\[ \sin x = 1 \quad \text{and} \quad \cos 2y = 1 \]
The solutions for \(\sin x = 1\) occur at:
\[ x = \frac{\pi}{2} + 2k\pi \quad (k \in \mathbb{Z}) \]

The solutions for \(\cos 2y = 1\) occur at:
\[ 2y = 2k\pi \implies y = k\pi \quad (k \in \mathbb{Z}) \]
Thus, within the interval \([0, 2\pi]\), the values of \(y\) are:
\[ y = 0, \pi \]

7. **Analyzing Equation (6):**
From equation (6):
\[ \cos x \sin 2y = 0 \]
This implies either:
\[ \cos x = 0 \quad \text{or} \quad \sin 2y = 0 \]
The solutions for \(\cos x = 0\) occur at:
\[ x = \frac{\pi}{2} + k\pi \quad (k \in \mathbb{Z}) \]
The solutions for \(\sin 2y = 0\) occur at:
\[ 2y = k\pi \implies y = \frac{k\pi}{2} \quad (k \in \mathbb{Z}) \]
Within the interval \([0, 2\pi]\), the values of \(y\) are:
\[ y = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi \]

8. **Conclusion:**
The permissible values of \(y\) from both equations are \(y = 0, \pi\) from equation (5) and \(y = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi\) from equation (6). The unique values of \(y\) in the interval \([0, 2\pi]\) are:
\[ y = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi \]
Therefore, the total number of distinct values of \(y\) is **5**.