The sum of value of `alpha, 0 le alpha le 2pi`, such that `x^(2) -2x cos alpha +1=0` has real roots is ……….. Times `pi`.

Answer

To solve the problem, we need to determine the values of \( \alpha \) for which the quadratic equation \( x^2 - 2x \cos \alpha + 1 = 0 \) has real roots.

### Step-by-Step Solution:

1. **Identify the quadratic equation**:
The given quadratic equation is:
\[ x^2 - 2x \cos \alpha + 1 = 0 \]

2. **Use the discriminant condition**:
For a quadratic equation \( ax^2 + bx + c = 0 \) to have real roots, its discriminant \( D \) must be non-negative:
\[ D = b^2 - 4ac \geq 0 \]
Here, \( a = 1 \), \( b = -2 \cos \alpha \), and \( c = 1 \).

3. **Calculate the discriminant**:
\[ D = (-2 \cos \alpha)^2 - 4 \cdot 1 \cdot 1 = 4 \cos^2 \alpha - 4 \]
Thus, we need:
\[ 4 \cos^2 \alpha - 4 \geq 0 \]

4. **Simplify the inequality**:
Dividing the entire inequality by 4:
\[ \cos^2 \alpha - 1 \geq 0 \]
This can be rewritten as:
\[ \cos^2 \alpha \geq 1 \]

5. **Solve for \( \cos \alpha \)**:
The inequality \( \cos^2 \alpha \geq 1 \) implies:
\[ \cos \alpha = \pm 1 \]

6. **Find the values of \( \alpha \)**:
- \( \cos \alpha = 1 \) occurs at:
\[ \alpha = 0 \quad \text{and} \quad \alpha = 2\pi \]
- \( \cos \alpha = -1 \) occurs at:
\[ \alpha = \pi \]

7. **List the valid values of \( \alpha \)**:
The values of \( \alpha \) that satisfy the condition are:
\[ \alpha = 0, \pi, 2\pi \]

8. **Calculate the sum of the values**:
Now, we sum these values:
\[ 0 + \pi + 2\pi = 3\pi \]

9. **Final answer**:
The sum of the values of \( \alpha \) such that the quadratic has real roots is:
\[ 3\pi \]

### Conclusion:
The sum of the values of \( \alpha \) is \( 3 \) times \( \pi \).

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