If the quadratic equation ,
`4^(sec^(2)alpha) x^(2) + 2x + (beta^(2)- beta + 1/2) = 0` have real roots, then find all the possible value of `cos alpha + cos^(-1) beta`.

Answer

To solve the problem, we need to analyze the given quadratic equation and determine the conditions under which it has real roots. The equation is:

\[ 4^{\sec^2 \alpha} x^2 + 2x + \left( \beta^2 - \beta + \frac{1}{2} \right) = 0 \]

### Step 1: Identify coefficients
In the standard form of a quadratic equation \( ax^2 + bx + c = 0 \), we can identify:
- \( a = 4^{\sec^2 \alpha} \)
- \( b = 2 \)
- \( c = \beta^2 - \beta + \frac{1}{2} \)

### Step 2: Use the discriminant condition
For the quadratic equation to have real roots, the discriminant \( D \) must be greater than or equal to zero:

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

Substituting the identified coefficients:

\[ D = 2^2 - 4 \cdot 4^{\sec^2 \alpha} \cdot \left( \beta^2 - \beta + \frac{1}{2} \right) \geq 0 \]

This simplifies to:

\[ 4 - 4 \cdot 4^{\sec^2 \alpha} \cdot \left( \beta^2 - \beta + \frac{1}{2} \right) \geq 0 \]

### Step 3: Rearranging the inequality
Rearranging the inequality gives:

\[ 1 \geq 4^{\sec^2 \alpha} \cdot \left( \beta^2 - \beta + \frac{1}{2} \right) \]

This can be rewritten as:

\[ \beta^2 - \beta + \frac{1}{2} \leq \frac{1}{4^{\sec^2 \alpha}} \]

### Step 4: Analyze the quadratic function
The expression \( \beta^2 - \beta + \frac{1}{2} \) is a quadratic function in \( \beta \). To find its minimum value, we can complete the square or use the vertex formula:

The vertex of the quadratic \( ax^2 + bx + c \) occurs at \( \beta = -\frac{b}{2a} \):

\[ \beta = \frac{1}{2} \]

Substituting \( \beta = \frac{1}{2} \) into the quadratic gives:

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

Thus, the minimum value of \( \beta^2 - \beta + \frac{1}{2} \) is \( \frac{1}{4} \).

### Step 5: Set up the inequality
Now we can set up the inequality:

\[ \frac{1}{4} \leq \frac{1}{4^{\sec^2 \alpha}} \]

This implies:

\[ 4^{\sec^2 \alpha} \leq 4 \]

Taking logarithm base 4 on both sides:

\[ \sec^2 \alpha \leq 1 \]

Since \( \sec^2 \alpha \geq 1 \) for all \( \alpha \), we conclude that:

\[ \sec^2 \alpha = 1 \implies \alpha = n\pi \quad (n \in \mathbb{Z}) \]

### Step 6: Find values of \( \cos \alpha + \cos^{-1} \beta \)
Since \( \sec^2 \alpha = 1 \), we have \( \cos \alpha = 1 \) or \( \cos \alpha = -1 \).

1. If \( \cos \alpha = 1 \):
\[ \cos \alpha + \cos^{-1} \beta = 1 + \cos^{-1} \left( \frac{1}{2} \right) = 1 + \frac{\pi}{3} = \frac{3}{3} + \frac{\pi}{3} = \frac{3 + \pi}{3} \]

2. If \( \cos \alpha = -1 \):
\[ \cos \alpha + \cos^{-1} \beta = -1 + \cos^{-1} \left( \frac{1}{2} \right) = -1 + \frac{\pi}{3} = \frac{-3 + \pi}{3} \]

### Final Answer
Thus, the possible values of \( \cos \alpha + \cos^{-1} \beta \) are:

\[ \frac{3 + \pi}{3} \quad \text{and} \quad \frac{-3 + \pi}{3} \]