The number of values of the parameter `alpha in [0, 2pi]` for which the quadratic function `(sin alpha)x^(2)+(2cosalpha)x+(1)/(2)(cos alpha+sinalpha)` is the square of a linear funcion is

Options

A. 2
B. 4
C. 6
D. 8

Answer

To solve the problem, we need to determine the number of values of the parameter \( \alpha \) in the interval \([0, 2\pi]\) for which the quadratic function

\[ f(x) = (\sin \alpha)x^2 + (2\cos \alpha)x + \frac{1}{2}(\cos \alpha + \sin \alpha) \]

is a perfect square of a linear function.

### Step 1: Identify the condition for a quadratic to be a perfect square

A quadratic function \( ax^2 + bx + c \) is a perfect square if it can be expressed in the form \( (px + q)^2 \). This means that the discriminant must be zero:

\[ D = b^2 - 4ac = 0 \]

### Step 2: Calculate the coefficients

For our quadratic function, we identify:

- \( a = \sin \alpha \)
- \( b = 2\cos \alpha \)
- \( c = \frac{1}{2}(\cos \alpha + \sin \alpha) \)

### Step 3: Set up the discriminant condition

The discriminant \( D \) can be calculated as follows:

\[ D = (2\cos \alpha)^2 - 4(\sin \alpha)\left(\frac{1}{2}(\cos \alpha + \sin \alpha)\right) \]

Calculating this gives:

\[ D = 4\cos^2 \alpha - 2\sin \alpha(\cos \alpha + \sin \alpha) \]

### Step 4: Simplify the discriminant

Expanding the second term:

\[ D = 4\cos^2 \alpha - 2\sin \alpha \cos \alpha - 2\sin^2 \alpha \]

Rearranging gives:

\[ D = 4\cos^2 \alpha - 2\sin \alpha \cos \alpha - 2\sin^2 \alpha \]

### Step 5: Set the discriminant to zero

Setting \( D = 0 \):

\[ 4\cos^2 \alpha - 2\sin \alpha \cos \alpha - 2\sin^2 \alpha = 0 \]

### Step 6: Factor the equation

This can be rearranged as:

\[ 2\cos^2 \alpha - \sin \alpha \cos \alpha - \sin^2 \alpha = 0 \]

### Step 7: Solve for \( \cot \alpha \)

Dividing through by \( \sin^2 \alpha \) (assuming \( \sin \alpha \neq 0 \)) gives:

\[ 2\cot^2 \alpha - \cot \alpha - 1 = 0 \]

Letting \( y = \cot \alpha \), we have:

\[ 2y^2 - y - 1 = 0 \]

### Step 8: Use the quadratic formula

Using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

\[ y = \frac{1 \pm \sqrt{1^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2} = \frac{1 \pm \sqrt{1 + 8}}{4} = \frac{1 \pm 3}{4} \]

This gives:

\[ y = 1 \quad \text{or} \quad y = -\frac{1}{2} \]

### Step 9: Find \( \alpha \) values

1. For \( \cot \alpha = 1 \):
- \( \alpha = \frac{\pi}{4}, \frac{5\pi}{4} \)

2. For \( \cot \alpha = -\frac{1}{2} \):
- \( \alpha = \tan^{-1}(-2) \) gives two angles in \([0, 2\pi]\).

### Step 10: Count the solutions

Thus, we have:

- 2 solutions from \( \cot \alpha = 1 \)
- 2 solutions from \( \cot \alpha = -\frac{1}{2} \)

### Conclusion

The total number of values of \( \alpha \) in the interval \([0, 2\pi]\) for which the quadratic function is a perfect square is:

\[ \boxed{4} \]