The equation `(cos p-1) x^(2) + cos p*x + sin p = 0` where x is a variable, has real roots. Then the interval of possible values of p is

Options

A. `(0, 2pi)`
B. `(-pi, 0)`
C. `(-pi/2, pi/2)`
D. `(0, pi)`

Answer

To determine the interval of possible values of \( p \) for which the equation

\[ (\cos p - 1)x^2 + \cos p \cdot x + \sin p = 0 \]

has real roots, we need to analyze the discriminant of the quadratic equation.

### Step-by-Step Solution:

1. **Identify the coefficients**:
The given quadratic equation can be expressed in the standard form \( ax^2 + bx + c = 0 \), where:
- \( a = \cos p - 1 \)
- \( b = \cos p \)
- \( c = \sin p \)

2. **Calculate the discriminant**:
The discriminant \( D \) of a quadratic equation \( ax^2 + bx + c = 0 \) is given by the formula:
\[ D = b^2 - 4ac \]
Substituting the values of \( a \), \( b \), and \( c \):
\[ D = (\cos p)^2 - 4(\cos p - 1)(\sin p) \]

3. **Simplify the discriminant**:
Expanding the expression:
\[ D = \cos^2 p - 4(\cos p \sin p - \sin p) \]
\[ = \cos^2 p - 4\cos p \sin p + 4\sin p \]

4. **Set the discriminant to be non-negative**:
For the quadratic equation to have real roots, the discriminant must be non-negative:
\[ D \geq 0 \]
This leads to the inequality:
\[ \cos^2 p - 4\cos p \sin p + 4\sin p \geq 0 \]

5. **Analyze the terms**:
Since \( \cos^2 p \geq 0 \) for all \( p \) and \( \cos p - 1 \leq 0 \) (as \( \cos p \) ranges from -1 to 1), we can conclude:
- \( \sin p \) must be non-negative for the entire expression to be non-negative.

6. **Determine the intervals for \( \sin p \)**:
The sine function is non-negative in the intervals:
\[ p \in [0, \pi] \]
This means that \( \sin p \geq 0 \) for \( p \) in this range.

7. **Conclusion**:
Therefore, the interval of possible values of \( p \) for which the given equation has real roots is:
\[ p \in [0, \pi] \]