Find the fundamental period of `f(x)=cosxcos2xcos3xdot`

Explanation

To find the fundamental period of the function \( f(x) = \cos x \cdot \cos 2x \cdot \cos 3x \), we will follow these steps:

### Step 1: Identify the periods of the individual cosine functions
The period of \( \cos x \) is \( 2\pi \).

For \( \cos 2x \):
\[ \text{Period} = \frac{2\pi}{2} = \pi \]

For \( \cos 3x \):
\[ \text{Period} = \frac{2\pi}{3} \]

### Step 2: Find the least common multiple (LCM) of the periods
We have the periods:
- \( 2\pi \)
- \( \pi \)
- \( \frac{2\pi}{3} \)

To find the LCM, we express these periods in a common form:
- \( 2\pi = \frac{6\pi}{3} \)
- \( \pi = \frac{3\pi}{3} \)
- \( \frac{2\pi}{3} = \frac{2\pi}{3} \)

The LCM of \( \frac{6\pi}{3}, \frac{3\pi}{3}, \frac{2\pi}{3} \) is \( \frac{6\pi}{3} = 2\pi \).

### Step 3: Verify if \( 2\pi \) is the fundamental period
We need to check if \( f(x) = f(x + 2\pi) \):
\[ f(x + 2\pi) = \cos(x + 2\pi) \cdot \cos(2(x + 2\pi)) \cdot \cos(3(x + 2\pi)) \]
Using the periodic property of cosine:
\[ \cos(x + 2\pi) = \cos x, \quad \cos(2(x + 2\pi)) = \cos(2x), \quad \cos(3(x + 2\pi)) = \cos(3x) \]
Thus,
\[ f(x + 2\pi) = \cos x \cdot \cos 2x \cdot \cos 3x = f(x) \]

### Step 4: Check for smaller periods
Next, we check if \( \pi \) is a period:
\[ f(x + \pi) = \cos(x + \pi) \cdot \cos(2(x + \pi)) \cdot \cos(3(x + \pi)) \]
Using the periodic properties:
\[ \cos(x + \pi) = -\cos x, \quad \cos(2(x + \pi)) = -\cos 2x, \quad \cos(3(x + \pi)) = -\cos 3x \]
Thus,
\[ f(x + \pi) = (-\cos x) \cdot (-\cos 2x) \cdot (-\cos 3x) = -\cos x \cdot \cos 2x \cdot \cos 3x \neq f(x) \]

### Conclusion
Since \( f(x + \pi) \neq f(x) \), \( \pi \) is not a period. Therefore, the fundamental period of \( f(x) \) is:
\[ \text{Fundamental Period} = 2\pi \]