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. 2
B. 3
C. 4
D. 5

Explanation

To solve the given system of equations:

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

2. **Analyzing the First Equation:**
The left-hand side (LHS) of the first equation, \(\sin x \cos 2y\), must be less than or equal to 1 because both \(\sin x\) and \(\cos 2y\) are bounded between -1 and 1. Therefore, we have:
\[ (a^2 - 1)^2 + 1 \leq 1 \]
This implies:
\[ (a^2 - 1)^2 \leq 0 \]
Since the square of any real number is non-negative, the only solution to this inequality is:
\[ (a^2 - 1)^2 = 0 \implies a^2 - 1 = 0 \implies a^2 = 1 \]
Thus, \(a = 1\) or \(a = -1\).

3. **Checking the Values of \(a\):**
- If \(a = 1\):
\[ \cos x \sin 2y = 1 + 1 = 2 \]
This is not possible since \(\cos x \sin 2y\) cannot exceed 1.
- If \(a = -1\):
\[ \cos x \sin 2y = -1 + 1 = 0 \]

4. **Substituting \(a = -1\) into the Equations:**
- The first equation becomes:
\[ \sin x \cos 2y = (0)^2 + 1 = 1 \]
- The second equation becomes:
\[ \cos x \sin 2y = 0 \]

5. **Solving the Second Equation:**
From \(\cos x \sin 2y = 0\), we have two cases:
- Case 1: \(\cos x = 0\) implies \(x = \frac{\pi}{2} + n\pi\) for \(n \in \mathbb{Z}\).
- Case 2: \(\sin 2y = 0\) implies \(2y = n\pi\) for \(n \in \mathbb{Z}\).

6. **Finding Values of \(y\):**
From \(2y = n\pi\):
- For \(n = 0\): \(y = 0\)
- For \(n = 1\): \(y = \frac{\pi}{2}\)
- For \(n = 2\): \(y = \pi\)
- For \(n = 3\): \(y = \frac{3\pi}{2}\)
- For \(n = 4\): \(y = 2\pi\)

Thus, the values of \(y\) in the interval \([0, 2\pi]\) are:
\[ y = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi \]

7. **Counting Distinct Values of \(y\):**
The distinct values of \(y\) are:
- \(0\)
- \(\frac{\pi}{2}\)
- \(\pi\)
- \(\frac{3\pi}{2}\)
- \(2\pi\)

Therefore, there are **5 distinct values of \(y\)**.

**Final Answer:**
The number of values of \(y\) in the interval \([0, 2\pi]\) when the system has a solution for permissible values of \(a\) is **5**.

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