To determine where the function \( f(x) = \cos\left(\frac{\pi}{[x]}\right) \cos\left(\frac{\pi}{2}(x-1)\right) \) is continuous, we will analyze the function step by step.
### Step 1: Understanding the Function
The function \( f(x) \) involves the greatest integer function, denoted as \( [x] \). This function returns the largest integer less than or equal to \( x \). Therefore, we need to consider the behavior of \( f(x) \) at integer and non-integer values of \( x \).
### Step 2: Check Continuity at \( x = 0 \)
At \( x = 0 \):
- \( [0] = 0 \)
- The term \( \frac{\pi}{[x]} \) becomes undefined since we cannot divide by zero. Hence, \( f(0) \) is not defined.
**Conclusion**: \( f(x) \) is not continuous at \( x = 0 \).
### Step 3: Check Continuity at \( x = 1 \)
At \( x = 1 \):
- \( [1] = 1 \)
- Therefore, \( f(1) = \cos(\pi) \cos(0) = -1 \cdot 1 = -1 \).
Now, we need to check the left-hand limit and right-hand limit at \( x = 1 \).
**Left-hand limit** (\( x \to 1^- \)):
- For \( x \) approaching \( 1 \) from the left, \( [x] = 0 \).
- Thus, \( f(1^-) = \cos(0) \cos\left(\frac{\pi}{2}(1-1)\right) = 1 \cdot 1 = 1 \).
**Right-hand limit** (\( x \to 1^+ \)):
- For \( x \) approaching \( 1 \) from the right, \( [x] = 1 \).
- Thus, \( f(1^+) = \cos(\pi) \cos\left(\frac{\pi}{2}(1-1)\right) = -1 \cdot 1 = -1 \).
Since \( f(1^-) \neq f(1) \) and \( f(1^+) \neq f(1) \), \( f(x) \) is not continuous at \( x = 1 \).
### Step 4: Check Continuity at \( x = 2 \)
At \( x = 2 \):
- \( [2] = 2 \)
- Therefore, \( f(2) = \cos\left(\frac{\pi}{2}\right) \cos(1) = 0 \cdot \cos(1) = 0 \).
**Left-hand limit** (\( x \to 2^- \)):
- For \( x \) approaching \( 2 \) from the left, \( [x] = 1 \).
- Thus, \( f(2^-) = \cos\left(\frac{\pi}{1}\right) \cos\left(\frac{\pi}{2}(2-1)\right) = -1 \cdot 1 = -1 \).
**Right-hand limit** (\( x \to 2^+ \)):
- For \( x \) approaching \( 2 \) from the right, \( [x] = 2 \).
- Thus, \( f(2^+) = \cos\left(\frac{\pi}{2}\right) \cos(1) = 0 \cdot \cos(1) = 0 \).
Since \( f(2^-) \neq f(2) \), \( f(x) \) is not continuous at \( x = 2 \).
### Step 5: Check Continuity at \( x = 3 \)
At \( x = 3 \):
- \( [3] = 3 \)
- Therefore, \( f(3) = \cos\left(\frac{\pi}{3}\right) \cos(2) = \frac{1}{2} \cos(2) \).
**Left-hand limit** (\( x \to 3^- \)):
- For \( x \) approaching \( 3 \) from the left, \( [x] = 2 \).
- Thus, \( f(3^-) = \cos\left(\frac{\pi}{2}\right) \cos(2) = 0 \cdot \cos(2) = 0 \).
**Right-hand limit** (\( x \to 3^+ \)):
- For \( x \) approaching \( 3 \) from the right, \( [x] = 3 \).
- Thus, \( f(3^+) = \cos\left(\frac{\pi}{3}\right) \cos(2) = \frac{1}{2} \cos(2) \).
Since \( f(3^-) \neq f(3) \), \( f(x) \) is not continuous at \( x = 3 \).
### Conclusion
After checking the continuity at \( x = 0, 1, 2, \) and \( 3 \), we find that \( f(x) \) is only continuous at integer values \( x = 1, 2, 3 \).
### Final Answer
The function \( f(x) \) is continuous at \( x = 1, 2, 3 \).
