Solution of the equation `cos^(2)x+cos^(2)2x+cos^(2)3x=1` is

Options

A. `x=(2m+1)(pi)/2(m in I)`
B. `x=(2n+1)(pi)/4(n in I)`
C. `x=(2k+1)(pi)/6(k in I)`
D. `x=(2l+1)(pi)/8(l in I)`

Explanation

To solve the equation \( \cos^2 x + \cos^2 2x + \cos^2 3x = 1 \), we can follow these steps:

### Step 1: Use the identity for cosine squared
We know that \( \cos^2 \theta = \frac{1 + \cos 2\theta}{2} \). We can apply this identity to each term in the equation.

### Step 2: Rewrite the equation
Using the identity, we can rewrite each term:
\[ \cos^2 x = \frac{1 + \cos 2x}{2}, \quad \cos^2 2x = \frac{1 + \cos 4x}{2}, \quad \cos^2 3x = \frac{1 + \cos 6x}{2} \]
Substituting these into the equation gives:
\[ \frac{1 + \cos 2x}{2} + \frac{1 + \cos 4x}{2} + \frac{1 + \cos 6x}{2} = 1 \]

### Step 3: Combine the fractions
Combining the fractions, we have:
\[ \frac{3 + \cos 2x + \cos 4x + \cos 6x}{2} = 1 \]

### Step 4: Multiply through by 2
To eliminate the fraction, multiply both sides by 2:
\[ 3 + \cos 2x + \cos 4x + \cos 6x = 2 \]

### Step 5: Simplify the equation
Rearranging gives:
\[ \cos 2x + \cos 4x + \cos 6x = -1 \]

### Step 6: Analyze the cosine terms
The maximum value of \( \cos \theta \) is 1, hence the sum \( \cos 2x + \cos 4x + \cos 6x \) can range from -3 to 3. The only way for this sum to equal -1 is if each cosine term takes on specific values.

### Step 7: Set up conditions
To satisfy \( \cos 2x + \cos 4x + \cos 6x = -1 \), we can consider cases where:
- Each cosine term is negative.
- The angles correspond to specific values where the cosine function is negative.

### Step 8: Solve for \( x \)
We can set \( \cos 2x = -1 \), \( \cos 4x = -1 \), and \( \cos 6x = -1 \) to find specific solutions:
- \( 2x = (2n + 1)\pi \) for integers \( n \)
- \( 4x = (2m + 1)\pi \) for integers \( m \)
- \( 6x = (2k + 1)\pi \) for integers \( k \)

From these, we can derive:
- \( x = \frac{(2n + 1)\pi}{2} \)
- \( x = \frac{(2m + 1)\pi}{4} \)
- \( x = \frac{(2k + 1)\pi}{6} \)

### Final Solution
The solutions for \( x \) can be expressed as:
\[ x = \frac{(2n + 1)\pi}{2}, \quad x = \frac{(2m + 1)\pi}{4}, \quad x = \frac{(2k + 1)\pi}{6} \]
where \( n, m, k \) are integers.