The number of non-negative integral vlaues of `n, n le 10` so that a root of the equation `n ^(2) sin ^(2)x -2 sin x - (2n+1)=0` lies in interval `[0, (pi)/(2)]` is:

Answer

To solve the problem, we need to find the number of non-negative integral values of \( n \) such that \( n \leq 10 \) and a root of the equation

\[ n^2 \sin^2 x - 2 \sin x - (2n + 1) = 0 \]

lies in the interval \([0, \frac{\pi}{2}]\).

### Step 1: Rewrite the equation
We can rewrite the equation in terms of \( t = \sin x \):

\[ n^2 t^2 - 2t - (2n + 1) = 0 \]

### Step 2: Identify coefficients
In this quadratic equation, the coefficients are:
- \( a = n^2 \)
- \( b = -2 \)
- \( c = -(2n + 1) \)

### Step 3: Apply the quadratic formula
Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

\[ t = \frac{2 \pm \sqrt{(-2)^2 - 4(n^2)(-(2n + 1))}}{2n^2} \]

### Step 4: Simplify the discriminant
Calculating the discriminant:

\[ b^2 - 4ac = 4 + 8n^3 + 4n^2 = 4(1 + 2n^2 + 2n^3) \]

### Step 5: Substitute back into the formula
Now substituting back into the formula gives:

\[ t = \frac{2 \pm 2\sqrt{1 + 2n^2 + 2n^3}}{2n^2} = \frac{1 \pm \sqrt{1 + 2n^2 + 2n^3}}{n^2} \]

### Step 6: Determine the range for \( t \)
For \( t = \sin x \) to lie in the interval \([0, 1]\), we need:

\[ 0 \leq \frac{1 \pm \sqrt{1 + 2n^2 + 2n^3}}{n^2} \leq 1 \]

### Step 7: Analyze the two cases
1. **Case 1:** Using the positive sign:
\[ 0 \leq \frac{1 + \sqrt{1 + 2n^2 + 2n^3}}{n^2} \leq 1 \]
This is always true for \( n \geq 0 \).

2. **Case 2:** Using the negative sign:
\[ 0 \leq \frac{1 - \sqrt{1 + 2n^2 + 2n^3}}{n^2} \leq 1 \]
This leads to:
\[ 1 - \sqrt{1 + 2n^2 + 2n^3} \geq 0 \implies \sqrt{1 + 2n^2 + 2n^3} \leq 1 \]
Squaring both sides gives:
\[ 1 + 2n^2 + 2n^3 \leq 1 \implies 2n^2 + 2n^3 \leq 0 \]
Thus, \( n(n + 1) \leq 0 \). This implies \( n \) must be \( 0 \).

### Step 8: Find valid values of \( n \)
Now we need to find the non-negative integral values of \( n \) such that \( n \leq 10 \) and \( n \geq 3 \) (from the earlier analysis). The valid values of \( n \) are \( 3, 4, 5, 6, 7, 8, 9, 10 \).

### Step 9: Count the values
The valid values of \( n \) are \( 3, 4, 5, 6, 7, 8, 9, 10 \), which gives us a total of:

\[ 10 - 3 + 1 = 8 \]

### Final Answer
The number of non-negative integral values of \( n \) such that a root of the equation lies in the interval \([0, \frac{\pi}{2}]\) is **8**.

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