The solution of the system of equations `x+(cosalpha)y=1 and (cosalpha)x+4y=2` satisfy `x ge(4)/(5)` and `yle(1)/(2)`, then the value of `alpha` can lie in the inverval

Options

A. `alpha in [(pi)/(4),(pi)/(3)]`
B. `alpha in [0,(pi)/(6)]`
C. `alpha in [(pi)/(6),(pi)/(3)]`
D. `alpha in [(pi)/(3),(pi)/(2)]`

Answer

To solve the system of equations given by:

1. \( x + (\cos \alpha)y = 1 \)
2. \( (\cos \alpha)x + 4y = 2 \)

we will use Cramer's rule to find the values of \( x \) and \( y \) in terms of \( \alpha \).

### Step 1: Write the Coefficient Matrix
The coefficient matrix \( A \) is given by:
\[ A = \begin{pmatrix} 1 & \cos \alpha \\ \cos \alpha & 4 \end{pmatrix} \]

### Step 2: Calculate the Determinant of \( A \)
The determinant \( \Delta \) of the matrix \( A \) is calculated as:
\[ \Delta = 1 \cdot 4 - (\cos \alpha)(\cos \alpha) = 4 - \cos^2 \alpha \]

### Step 3: Calculate \( \Delta_1 \) and \( \Delta_2 \)
Now we calculate \( \Delta_1 \) and \( \Delta_2 \):

- For \( \Delta_1 \), replace the first column with the right-hand side values:
\[ \Delta_1 = \begin{vmatrix} 1 & \cos \alpha \\ 2 & 4 \end{vmatrix} = 1 \cdot 4 - 2 \cdot \cos \alpha = 4 - 2 \cos \alpha \]

- For \( \Delta_2 \), replace the second column with the right-hand side values:
\[ \Delta_2 = \begin{vmatrix} 1 & 1 \\ \cos \alpha & 2 \end{vmatrix} = 1 \cdot 2 - 1 \cdot \cos \alpha = 2 - \cos \alpha \]

### Step 4: Find \( x \) and \( y \)
Using Cramer's rule:
\[ x = \frac{\Delta_1}{\Delta} = \frac{4 - 2 \cos \alpha}{4 - \cos^2 \alpha} \]
\[ y = \frac{\Delta_2}{\Delta} = \frac{2 - \cos \alpha}{4 - \cos^2 \alpha} \]

### Step 5: Apply the Given Conditions
We have the conditions:
1. \( x \geq \frac{4}{5} \)
2. \( y \leq \frac{1}{2} \)

#### Condition 1: \( x \geq \frac{4}{5} \)
\[ \frac{4 - 2 \cos \alpha}{4 - \cos^2 \alpha} \geq \frac{4}{5} \]
Cross-multiplying gives:
\[ 5(4 - 2 \cos \alpha) \geq 4(4 - \cos^2 \alpha) \]
Expanding both sides:
\[ 20 - 10 \cos \alpha \geq 16 - 4 \cos^2 \alpha \]
Rearranging gives:
\[ 4 \cos^2 \alpha - 10 \cos \alpha + 4 \geq 0 \]

#### Condition 2: \( y \leq \frac{1}{2} \)
\[ \frac{2 - \cos \alpha}{4 - \cos^2 \alpha} \leq \frac{1}{2} \]
Cross-multiplying gives:
\[ 2(2 - \cos \alpha) \leq 4 - \cos^2 \alpha \]
Expanding both sides:
\[ 4 - 2 \cos \alpha \leq 4 - \cos^2 \alpha \]
Rearranging gives:
\[ \cos^2 \alpha - 2 \cos \alpha \geq 0 \]
Factoring gives:
\[ \cos \alpha (\cos \alpha - 2) \geq 0 \]

### Step 6: Solve the Inequalities
1. From \( 4 \cos^2 \alpha - 10 \cos \alpha + 4 \geq 0 \), use the quadratic formula to find the roots:
\[ \cos \alpha = \frac{10 \pm \sqrt{100 - 64}}{8} = \frac{10 \pm 6}{8} \]
This gives \( \cos \alpha = 2 \) and \( \cos \alpha = \frac{1}{2} \).

2. From \( \cos \alpha (\cos \alpha - 2) \geq 0 \), the solution is \( \cos \alpha \leq 0 \) or \( \cos \alpha \geq 2 \) (not possible since \( \cos \alpha \) cannot exceed 1).

### Step 7: Determine the Interval for \( \alpha \)
The valid range for \( \alpha \) based on \( \cos \alpha \) is:
\[ \frac{1}{2} \leq \cos \alpha \leq 1 \]
This corresponds to the angles:
\[ \alpha \in [\frac{\pi}{3}, \frac{\pi}{2}] \]

### Final Answer
Thus, the value of \( \alpha \) can lie in the interval:
\[ \boxed{[\frac{\pi}{3}, \frac{\pi}{2}]} \]