If ` x^(2) - 2 x cos theta + 1 = 0 " then " x^(2 n) - 2 x ^(n) cos n theta ` is equal to

Options

A. 1
B. `-1`
C. 0
D. none of these

Answer

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

### Step 1: Solve the Quadratic Equation
The given quadratic equation is:
\[ x^2 - 2x \cos \theta + 1 = 0 \]
We can use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1 \), \( b = -2 \cos \theta \), and \( c = 1 \).

### Step 2: Calculate the Discriminant
First, we calculate the discriminant:
\[ b^2 - 4ac = (-2 \cos \theta)^2 - 4 \cdot 1 \cdot 1 = 4 \cos^2 \theta - 4 = 4(\cos^2 \theta - 1) = 4(-\sin^2 \theta) = -4 \sin^2 \theta \]

### Step 3: Find the Roots
Now substituting back into the quadratic formula:
\[ x = \frac{2 \cos \theta \pm \sqrt{-4 \sin^2 \theta}}{2} \]
This simplifies to:
\[ x = \cos \theta \pm i \sin \theta \]
Using Euler's formula, we can express the roots as:
\[ x_1 = e^{i\theta}, \quad x_2 = e^{-i\theta} \]

### Step 4: Calculate \( x^n \)
Now we need to find \( x^n \):
\[ x_1^n = (e^{i\theta})^n = e^{in\theta} = \cos n\theta + i \sin n\theta \]
\[ x_2^n = (e^{-i\theta})^n = e^{-in\theta} = \cos n\theta - i \sin n\theta \]

### Step 5: Calculate \( x^{2n} - 2x^n \cos n\theta \)
Now, we can calculate:
\[ x^{2n} = (x^n)^2 \]
For both roots:
\[ x_1^{2n} = (x_1^n)^2 = (\cos n\theta + i \sin n\theta)^2 = \cos^2 n\theta - \sin^2 n\theta + 2i \cos n\theta \sin n\theta \]
\[ x_2^{2n} = (x_2^n)^2 = (\cos n\theta - i \sin n\theta)^2 = \cos^2 n\theta - \sin^2 n\theta - 2i \cos n\theta \sin n\theta \]

### Step 6: Combine the Results
Now, we calculate:
\[ x^{2n} - 2x^n \cos n\theta = (x_1^{2n} + x_2^{2n}) - 2(\cos n\theta)(x_1^n + x_2^n) \]
Since \( x_1^n + x_2^n = 2\cos n\theta \):
\[ x^{2n} - 2x^n \cos n\theta = 2(\cos^2 n\theta - \sin^2 n\theta) - 4\cos^2 n\theta = -2\cos^2 n\theta - 2\sin^2 n\theta \]
Using the identity \( \cos^2 n\theta + \sin^2 n\theta = 1 \):
\[ x^{2n} - 2x^n \cos n\theta = -2 \]

### Final Answer
Thus, the expression \( x^{2n} - 2x^n \cos n\theta \) is equal to:
\[ \boxed{-2} \]