If there exists at least one real x which satisfies both the equatios `x^(2)+2xsiny+1=0`, where `yin(0,pi//2),andax^(2)+x+1=0`, then the value of `a+siny` is ______.

Explanation

To solve the problem, we need to find the value of \( a + \sin y \) given the two equations:

1. \( x^2 + 2x \sin y + 1 = 0 \)
2. \( ax^2 + x + 1 = 0 \)

### Step 1: Analyze the first equation
The first equation is a quadratic in \( x \). For it to have real roots, the discriminant must be non-negative.

The discriminant \( D_1 \) for the first equation \( x^2 + 2x \sin y + 1 = 0 \) is given by:

\[ D_1 = b^2 - 4ac \]

Here, \( a = 1 \), \( b = 2 \sin y \), and \( c = 1 \). Thus,

\[ D_1 = (2 \sin y)^2 - 4 \cdot 1 \cdot 1 = 4 \sin^2 y - 4 \]

For the roots to be real, we need:

\[ D_1 \geq 0 \implies 4 \sin^2 y - 4 \geq 0 \implies \sin^2 y \geq 1 \]

However, since \( \sin y \) is bounded between 0 and 1 for \( y \in (0, \frac{\pi}{2}) \), we conclude that \( D_1 < 0 \) unless \( \sin y = 1 \). Therefore, we must have \( \sin y = 1 \).

### Step 2: Substitute \( \sin y \)
Since \( \sin y = 1 \), we can substitute this value into the first equation:

\[ x^2 + 2x(1) + 1 = 0 \implies x^2 + 2x + 1 = 0 \implies (x + 1)^2 = 0 \]

This gives us a double root at \( x = -1 \).

### Step 3: Analyze the second equation
Next, we analyze the second equation \( ax^2 + x + 1 = 0 \). For this equation to have a common root with the first equation, the value \( x = -1 \) must also satisfy it.

Substituting \( x = -1 \) into the second equation:

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

### Step 4: Calculate \( a + \sin y \)
Now that we have \( a = 0 \) and \( \sin y = 1 \):

\[ a + \sin y = 0 + 1 = 1 \]

### Final Answer
Thus, the value of \( a + \sin y \) is:

\[ \boxed{1} \]