If `7theta=(2n+1)pi`, where `n=0,1,2,3,4,5,6`, then answer the following questions.
The equations whose roots are `cos. (pi)/(7), cos. (3pi)/(7), cos. (5pi)/(7)` is

Options

A. `8x^(2)+4x^(2)+4x+1=0`
B. `8x^(3)-4x^(2)-4x-1=0`
C. `8x^(3)-4x^(2)-4x-1=0`
D. `8x^(3)+4x^(2)+4x-1=0`

Answer

To solve the problem, we need to find the polynomial equation whose roots are \( \cos\left(\frac{\pi}{7}\right) \), \( \cos\left(\frac{3\pi}{7}\right) \), and \( \cos\left(\frac{5\pi}{7}\right) \).

### Step 1: Identify the roots
The roots we are dealing with are:
- \( x_1 = \cos\left(\frac{\pi}{7}\right) \)
- \( x_2 = \cos\left(\frac{3\pi}{7}\right) \)
- \( x_3 = \cos\left(\frac{5\pi}{7}\right) \)

### Step 2: Use the fact about roots of unity
The roots can be derived from the equation \( \cos(7\theta) = 0 \), which gives us \( 7\theta = (2n + 1)\pi \) for \( n = 0, 1, 2, 3, 4, 5, 6 \). This implies:
- \( \theta = \frac{(2n + 1)\pi}{7} \)

### Step 3: Find the polynomial
The polynomial whose roots are \( \cos\left(\frac{\pi}{7}\right) \), \( \cos\left(\frac{3\pi}{7}\right) \), and \( \cos\left(\frac{5\pi}{7}\right) \) can be derived from the Chebyshev polynomial of the first kind. The relevant polynomial is given by:
\[ T_7(x) = 64x^7 - 112x^5 + 56x^3 - 7x \]
However, we are interested in the polynomial whose roots are \( \cos\left(\frac{\pi}{7}\right) \), \( \cos\left(\frac{3\pi}{7}\right) \), and \( \cos\left(\frac{5\pi}{7}\right) \).

### Step 4: Factor the polynomial
The polynomial can be factored to find the quadratic polynomial whose roots are the cosines we need. The polynomial can be expressed as:
\[ 8x^3 - 4x^2 - 4x - 1 = 0 \]
This polynomial has the desired roots.

### Final Answer
Thus, the equation whose roots are \( \cos\left(\frac{\pi}{7}\right) \), \( \cos\left(\frac{3\pi}{7}\right) \), and \( \cos\left(\frac{5\pi}{7}\right) \) is:
\[ 8x^3 - 4x^2 - 4x - 1 = 0 \]