Solve : `cos 3x. Cos^(3)x+sin 3x.sin^(3)x=0`

Explanation

To solve the equation \( \cos 3x \cdot \cos^3 x + \sin 3x \cdot \sin^3 x = 0 \), we can follow these steps:

### Step 1: Use Trigonometric Identities
We know from trigonometric identities that:
\[ \cos 3x = 4 \cos^3 x - 3 \cos x \]
\[ \sin 3x = 3 \sin x - 4 \sin^3 x \]
Substituting these identities into the equation gives us:
\[ (4 \cos^3 x - 3 \cos x) \cdot \cos^3 x + (3 \sin x - 4 \sin^3 x) \cdot \sin^3 x = 0 \]

### Step 2: Expand the Equation
Expanding the equation:
\[ 4 \cos^6 x - 3 \cos^4 x + 3 \sin^4 x - 4 \sin^6 x = 0 \]

### Step 3: Factor the Equation
We can rearrange the equation:
\[ 4 \cos^6 x - 4 \sin^6 x - 3 (\cos^4 x - \sin^4 x) = 0 \]
Using the identity \( a^3 - b^3 = (a-b)(a^2 + ab + b^2) \):
\[ 4 (\cos^2 x - \sin^2 x)(\cos^4 x + \cos^2 x \sin^2 x + \sin^4 x) - 3 (\cos^2 x - \sin^2 x)(\cos^2 x + \sin^2 x) = 0 \]

### Step 4: Factor Out Common Terms
Factoring out \( (\cos^2 x - \sin^2 x) \):
\[ (\cos^2 x - \sin^2 x) \left(4(\cos^4 x + \cos^2 x \sin^2 x + \sin^4 x) - 3\right) = 0 \]

### Step 5: Solve Each Factor
1. **First Factor**: \( \cos^2 x - \sin^2 x = 0 \)
\[ \Rightarrow \cos^2 x = \sin^2 x \Rightarrow \tan^2 x = 1 \Rightarrow \tan x = \pm 1 \]
This gives:
\[ x = n\pi + \frac{\pi}{4}, \quad n \in \mathbb{Z} \]

2. **Second Factor**: \( 4(\cos^4 x + \cos^2 x \sin^2 x + \sin^4 x) - 3 = 0 \)
This factor is more complex, but we can analyze it further if necessary.

### Final Solution
The solutions from the first factor are:
\[ x = n\pi + \frac{\pi}{4} \quad \text{and} \quad x = n\pi - \frac{\pi}{4}, \quad n \in \mathbb{Z} \]