Find the period of the functions (where [ * ] denotes greatest integer function)
`f(x) = cos"" (3)/(5) x - sin""(2)/(7) x`.

Answer

To find the period of the function \( f(x) = \cos\left(\frac{3}{5} x\right) - \sin\left(\frac{2}{7} x\right) \), we will follow these steps:

### Step 1: Identify the periods of the individual trigonometric functions.

1. **For the cosine function**: The period of \( \cos(kx) \) is given by \( \frac{2\pi}{k} \).
- Here, \( k = \frac{3}{5} \).
- Therefore, the period of \( \cos\left(\frac{3}{5} x\right) \) is:
\[ T_1 = \frac{2\pi}{\frac{3}{5}} = \frac{2\pi \cdot 5}{3} = \frac{10\pi}{3} \]

2. **For the sine function**: The period of \( \sin(kx) \) is also given by \( \frac{2\pi}{k} \).
- Here, \( k = \frac{2}{7} \).
- Therefore, the period of \( \sin\left(\frac{2}{7} x\right) \) is:
\[ T_2 = \frac{2\pi}{\frac{2}{7}} = \frac{2\pi \cdot 7}{2} = 7\pi \]

### Step 2: Determine the least common multiple (LCM) of the two periods.

1. We have:
- \( T_1 = \frac{10\pi}{3} \)
- \( T_2 = 7\pi \)

2. To find the LCM, we first express both periods with a common denominator:
- Convert \( 7\pi \) to a fraction with a denominator of 3:
\[ 7\pi = \frac{21\pi}{3} \]

3. Now we have:
- \( T_1 = \frac{10\pi}{3} \)
- \( T_2 = \frac{21\pi}{3} \)

4. The LCM of the numerators (10 and 21) is:
- The prime factorization of 10 is \( 2 \times 5 \).
- The prime factorization of 21 is \( 3 \times 7 \).
- The LCM of 10 and 21 is \( 2 \times 3 \times 5 \times 7 = 210 \).

5. Therefore, the LCM of the periods is:
\[ \text{LCM} = \frac{210\pi}{3} = 70\pi \]

### Step 3: Conclusion

The period of the function \( f(x) = \cos\left(\frac{3}{5} x\right) - \sin\left(\frac{2}{7} x\right) \) is:
\[ \boxed{70\pi} \]