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 a for which the system has a solution is

Options

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

Explanation

To solve the system of equations

1. \( \sin x \cos 2y = (a^2 - 1)^2 + 1 \)
2. \( \cos x \sin 2y = a + 1 \)

we need to analyze the constraints on \( a \) that allow for solutions to these equations.

### Step 1: Analyze the first equation

The left-hand side of the first equation, \( \sin x \cos 2y \), can take values between -1 and 1, since both sine and cosine functions are bounded by these values. Therefore, we have:

\[ \sin x \cos 2y \leq 1 \]

This implies:

\[ (a^2 - 1)^2 + 1 \leq 1 \]

### Step 2: Simplify the inequality

From the inequality above, we can simplify:

\[ (a^2 - 1)^2 + 1 \leq 1 \implies (a^2 - 1)^2 \leq 0 \]

Since the square of any real number is non-negative, the only solution to this inequality is:

\[ (a^2 - 1)^2 = 0 \]

### Step 3: Solve for \( a \)

Taking the square root of both sides gives:

\[ a^2 - 1 = 0 \implies a^2 = 1 \implies a = 1 \text{ or } a = -1 \]

### Step 4: Check the second equation

Now we need to check if both values of \( a \) (i.e., \( a = 1 \) and \( a = -1 \)) satisfy the second equation:

1. For \( a = 1 \):
\[ \cos x \sin 2y = 1 + 1 = 2 \]
This is not possible since \( \cos x \) and \( \sin 2y \) can each only take values between -1 and 1.

2. For \( a = -1 \):
\[ \cos x \sin 2y = -1 + 1 = 0 \]
This is possible, as \( \cos x \sin 2y = 0 \) can be satisfied if either \( \cos x = 0 \) or \( \sin 2y = 0 \).

### Conclusion

The only value of \( a \) that allows the system of equations to have a solution is \( a = -1 \). Therefore, the number of values of \( a \) for which the system has a solution is:

\[ \boxed{1} \]