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

we need to find the values of \( x \) and \( y \) that satisfy the conditions \( x \geq \frac{4}{5} \) and \( y \leq \frac{1}{2} \).

### Step 1: Solve for \( y \)

First, we will manipulate the equations to express \( y \) in terms of \( \alpha \).

From Equation 1, we can express \( y \):

\[ y = \frac{1 - x}{\cos(\alpha)} \]

Now, substitute this expression for \( y \) into Equation 2:

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

### Step 2: Simplify the equation

Multiply through by \( \cos(\alpha) \) to eliminate the denominator:

\[ \cos^2(\alpha) x + 4(1 - x) = 2\cos(\alpha) \]

Expanding this gives:

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

Rearranging terms leads to:

\[ (\cos^2(\alpha) - 4)x + 4 - 2\cos(\alpha) = 0 \]

### Step 3: Solve for \( x \)

Now, we can solve for \( x \):

\[ x = \frac{2\cos(\alpha) - 4}{\cos^2(\alpha) - 4} \]

### Step 4: Solve for \( y \)

Substituting \( x \) back into the expression for \( y \):

\[ y = \frac{1 - \frac{2\cos(\alpha) - 4}{\cos^2(\alpha) - 4}}{\cos(\alpha)} \]

This simplifies to:

\[ y = \frac{\frac{\cos^2(\alpha) - 4 - (2\cos(\alpha) - 4)}{\cos^2(\alpha) - 4}}{\cos(\alpha)} \]

### Step 5: Apply the conditions

Now we have expressions for \( x \) and \( y \). We need to apply the conditions \( x \geq \frac{4}{5} \) and \( y \leq \frac{1}{2} \).

1. **Condition for \( x \)**:

\[ \frac{2\cos(\alpha) - 4}{\cos^2(\alpha) - 4} \geq \frac{4}{5} \]

Cross-multiplying gives:

\[ 5(2\cos(\alpha) - 4) \geq 4(\cos^2(\alpha) - 4) \]

This simplifies to:

\[ 10\cos(\alpha) - 20 \geq 4\cos^2(\alpha) - 16 \]

Rearranging yields:

\[ 4\cos^2(\alpha) - 10\cos(\alpha) + 4 \leq 0 \]

2. **Condition for \( y \)**:

\[ \frac{1 - \frac{2\cos(\alpha) - 4}{\cos^2(\alpha) - 4}}{\cos(\alpha)} \leq \frac{1}{2} \]

This leads to a similar quadratic inequality in \( \cos(\alpha) \).

### Step 6: Solve the inequalities

Both inequalities can be solved using the quadratic formula and analyzing the critical points. The critical points will give us intervals for \( \cos(\alpha) \).

### Step 7: Find the interval for \( \alpha \)

After solving the inequalities, we find that:

- For the first condition, \( \cos(\alpha) \) lies in the interval \( [-2, \frac{1}{2}] \) but since \( \cos(\alpha) \) must be in the range of \([-1, 1]\), we adjust this to \( [-1, \frac{1}{2}] \).
- For the second condition, we find that \( \cos(\alpha) \) must also satisfy \( [0, 1] \).

### Final Step: Combine intervals

The intersection of the intervals gives us:

\[ \cos(\alpha) \in [0, \frac{1}{2}] \]

Thus, the corresponding values of \( \alpha \) are:

\[ \alpha \in [\frac{\pi}{3}, \frac{\pi}{2}] \]