The solution of the equation
`9 cos^12 x + cos^2 2x + 1 = 6 cos ^6 x cos 2x. + 6 cos^6 x - 2 cos 2x`
are given by x =

Options

A. `n pi + pi/2`
B. `n pi + cos^(-1) "" sqrt(2/3)`
C. `n pi - cos^(-1) "" sqrt(2/3)`
D. none

Answer

To solve the equation \( 9 \cos^{12} x + \cos^2 2x + 1 = 6 \cos^6 x \cos 2x + 6 \cos^6 x - 2 \cos 2x \), we will follow these steps:

### Step 1: Substitute \( \cos 2x \)
Using the identity \( \cos 2x = 2 \cos^2 x - 1 \), we can rewrite the equation.

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

### Step 2: Expand the equation
Substituting \( \cos 2x \) into the equation gives:

\[ 9 \cos^{12} x + (2 \cos^2 x - 1)^2 + 1 = 6 \cos^6 x (2 \cos^2 x - 1) + 6 \cos^6 x - 2(2 \cos^2 x - 1) \]

### Step 3: Simplify the left side
Expanding \( (2 \cos^2 x - 1)^2 \):

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

Thus, the left side becomes:

\[ 9 \cos^{12} x + 4 \cos^4 x - 4 \cos^2 x + 1 + 1 = 9 \cos^{12} x + 4 \cos^4 x - 4 \cos^2 x + 2 \]

### Step 4: Simplify the right side
Now, simplifying the right side:

\[ 6 \cos^6 x (2 \cos^2 x - 1) + 6 \cos^6 x - 4 \cos^2 x + 2 \]

This simplifies to:

\[ 12 \cos^8 x - 6 \cos^6 x + 6 \cos^6 x - 4 \cos^2 x + 2 = 12 \cos^8 x - 4 \cos^2 x + 2 \]

### Step 5: Set the equation to zero
Now we have:

\[ 9 \cos^{12} x + 4 \cos^4 x - 4 \cos^2 x + 2 = 12 \cos^8 x - 4 \cos^2 x + 2 \]

Subtracting the right side from the left gives:

\[ 9 \cos^{12} x + 4 \cos^4 x - 12 \cos^8 x = 0 \]

### Step 6: Factor the equation
Factoring out \( \cos^4 x \):

\[ \cos^4 x (9 \cos^8 x - 12 \cos^4 x + 4) = 0 \]

### Step 7: Solve for \( \cos^4 x = 0 \)
This gives us:

1. \( \cos^4 x = 0 \) which implies \( \cos x = 0 \) leading to:
\[ x = n \frac{\pi}{2} + \frac{\pi}{2}, \quad n \in \mathbb{Z} \]

### Step 8: Solve the quadratic equation
Now, we solve the quadratic equation:

\[ 9 \cos^8 x - 12 \cos^4 x + 4 = 0 \]

Let \( y = \cos^4 x \):

\[ 9y^2 - 12y + 4 = 0 \]

### Step 9: Use the quadratic formula
Using the quadratic formula:

\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{12 \pm \sqrt{(-12)^2 - 4 \cdot 9 \cdot 4}}{2 \cdot 9} \]

Calculating the discriminant:

\[ = \frac{12 \pm \sqrt{144 - 144}}{18} = \frac{12}{18} = \frac{2}{3} \]

### Step 10: Solve for \( \cos^4 x \)
Thus, we have:

\[ \cos^4 x = \frac{2}{3} \]

Taking the fourth root gives:

\[ \cos x = \pm \left( \frac{2}{3} \right)^{1/4} \]

### Final Solution
Thus, the solutions for \( x \) are:

\[ x = n\pi \pm \cos^{-1}\left( \left( \frac{2}{3} \right)^{1/4} \right), \quad n \in \mathbb{Z} \]