To solve the equation \( \cos^{-1}(1-x) + m \cos^{-1}(x) = \frac{n\pi}{2} \), where \( m > 0 \) and \( n \leq 0 \), we will analyze the components of the equation step by step.
### Step 1: Understand the Domains and Ranges
1. The function \( \cos^{-1}(x) \) is defined for \( x \) in the range \([-1, 1]\) and its output (range) is \([0, \pi]\).
2. For \( \cos^{-1}(1-x) \) to be defined, \( 1-x \) must also lie within \([-1, 1]\). This implies:
\[
-1 \leq 1 - x \leq 1 \implies 0 \leq x \leq 2
\]
However, since \( x \) must also be in the domain of \( \cos^{-1}(x) \), we restrict \( x \) to \([0, 1]\).
### Step 2: Analyze the Ranges
1. The range of \( \cos^{-1}(1-x) \) as \( x \) varies from \( 0 \) to \( 1 \):
- When \( x = 0 \), \( \cos^{-1}(1-0) = \cos^{-1}(1) = 0 \).
- When \( x = 1 \), \( \cos^{-1}(1-1) = \cos^{-1}(0) = \frac{\pi}{2} \).
- Therefore, the range of \( \cos^{-1}(1-x) \) is \([0, \frac{\pi}{2}]\).
2. The range of \( m \cos^{-1}(x) \) as \( x \) varies from \( 0 \) to \( 1 \):
- When \( x = 0 \), \( m \cos^{-1}(0) = m \cdot \frac{\pi}{2} \).
- When \( x = 1 \), \( m \cos^{-1}(1) = m \cdot 0 = 0 \).
- Therefore, the range of \( m \cos^{-1}(x) \) is \([0, m \cdot \frac{\pi}{2}]\).
### Step 3: Combine the Ranges
1. The left-hand side of the equation \( \cos^{-1}(1-x) + m \cos^{-1}(x) \) will range from:
- Minimum: \( 0 + 0 = 0 \)
- Maximum: \( \frac{\pi}{2} + m \cdot \frac{\pi}{2} = \frac{(1+m)\pi}{2} \)
### Step 4: Analyze the Right-Hand Side
1. The right-hand side of the equation is \( \frac{n\pi}{2} \) where \( n \leq 0 \). This means:
- The right-hand side can take values \( 0, -\frac{\pi}{2}, -\pi, \ldots \) (negative or zero).
### Step 5: Determine the Number of Solutions
1. Since \( m > 0 \), the maximum value of the left-hand side \( \frac{(1+m)\pi}{2} \) is always positive.
2. The left-hand side can only take values in the range \([0, \frac{(1+m)\pi}{2}]\).
3. The right-hand side \( \frac{n\pi}{2} \) is non-positive (zero or negative).
4. Therefore, the left-hand side (which is always non-negative) cannot equal the right-hand side (which is non-positive).
### Conclusion
Since the left-hand side cannot equal the right-hand side, the number of solutions to the equation is:
\[
\boxed{0}
\]
