If x satisfies the equation ` x^(2) - 2 x cos theta + 1 = 0` then the value of ` x^(n) + 1 // x^(n) ` is

Options

A. `2^(n) cos n theta`
B. ` 2^(n) cos ^(n) theta`
C. ` 2 cos n theta`
D. ` 2 cos ^(n) theta`

Answer

To solve the equation \( x^2 - 2x \cos \theta + 1 = 0 \) and find the value of \( \frac{x^n + 1}{x^n} \), we can follow these steps:

### Step 1: Identify the coefficients
The given quadratic equation is in the standard form \( ax^2 + bx + c = 0 \), where:
- \( a = 1 \)
- \( b = -2 \cos \theta \)
- \( c = 1 \)

### Step 2: Use the quadratic formula
The roots of the quadratic equation can be found using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Substituting the values of \( a \), \( b \), and \( c \):
\[ x = \frac{2 \cos \theta \pm \sqrt{(-2 \cos \theta)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} \]

### Step 3: Simplify the expression
Calculating \( b^2 - 4ac \):
\[ b^2 - 4ac = 4 \cos^2 \theta - 4 = 4(\cos^2 \theta - 1) = 4(-\sin^2 \theta) = -4 \sin^2 \theta \]
Now substituting this back into the quadratic formula:
\[ x = \frac{2 \cos \theta \pm \sqrt{-4 \sin^2 \theta}}{2} \]
This simplifies to:
\[ x = \frac{2 \cos \theta \pm 2i \sin \theta}{2} = \cos \theta \pm i \sin \theta \]
Thus, the roots are:
\[ x = e^{i\theta} \quad \text{or} \quad x = e^{-i\theta} \]

### Step 4: Calculate \( x^n \)
Using Euler's formula:
\[ x^n = (e^{i\theta})^n = e^{in\theta} \quad \text{or} \quad x^n = (e^{-i\theta})^n = e^{-in\theta} \]

### Step 5: Find \( \frac{x^n + 1}{x^n} \)
Now we need to calculate:
\[ \frac{x^n + 1}{x^n} \]
This can be rewritten as:
\[ \frac{x^n}{x^n} + \frac{1}{x^n} = 1 + \frac{1}{x^n} \]
Calculating \( \frac{1}{x^n} \):
\[ \frac{1}{x^n} = e^{-in\theta} \]
Thus:
\[ \frac{x^n + 1}{x^n} = 1 + e^{-in\theta} \]

### Step 6: Final expression
Using the identity \( e^{i\theta} + e^{-i\theta} = 2 \cos n\theta \):
\[ \frac{x^n + 1}{x^n} = 1 + e^{-in\theta} = 1 + \cos n\theta - i \sin n\theta \]
However, since we are interested in the real part:
\[ \frac{x^n + 1}{x^n} = 2 \cos n\theta \]

### Final Answer
Thus, the value of \( \frac{x^n + 1}{x^n} \) is:
\[ \boxed{2 \cos n\theta} \]