The number of solution `(s)` of the equation `sin^-1x + cos^-1(1-x)=sin^-1(-х)` is/are

Explanation

To solve the equation \( \sin^{-1}x + \cos^{-1}(1-x) = \sin^{-1}(-x) \), we will follow these steps:

### Step 1: Rewrite the equation
The equation is given as:
\[ \sin^{-1}x + \cos^{-1}(1-x) = \sin^{-1}(-x) \]

### Step 2: Use the identity for inverse sine and cosine
We know that:
\[ \sin^{-1}(-x) = -\sin^{-1}(x) \]
Thus, we can rewrite the equation as:
\[ \sin^{-1}x + \cos^{-1}(1-x) = -\sin^{-1}(x) \]

### Step 3: Rearranging the equation
Rearranging gives:
\[ \sin^{-1}x + \sin^{-1}(x) + \cos^{-1}(1-x) = 0 \]
This simplifies to:
\[ 2\sin^{-1}x + \cos^{-1}(1-x) = 0 \]

### Step 4: Isolate \(\cos^{-1}(1-x)\)
Now, we can isolate \(\cos^{-1}(1-x)\):
\[ \cos^{-1}(1-x) = -2\sin^{-1}x \]

### Step 5: Use the cosine identity
Using the identity \( \cos^{-1}(y) = \pi/2 - \sin^{-1}(y) \), we can rewrite:
\[ \pi/2 - \sin^{-1}(1-x) = -2\sin^{-1}(x) \]
Rearranging gives:
\[ \sin^{-1}(1-x) = \pi/2 + 2\sin^{-1}(x) \]

### Step 6: Apply sine to both sides
Taking the sine of both sides, we have:
\[ 1-x = \sin\left(\frac{\pi}{2} + 2\sin^{-1}(x)\right) \]
Using the sine addition formula:
\[ \sin\left(\frac{\pi}{2} + \theta\right) = \cos(\theta) \]
Thus:
\[ 1-x = \cos(2\sin^{-1}(x)) \]

### Step 7: Use the double angle formula for cosine
Using the double angle formula:
\[ \cos(2\theta) = 1 - 2\sin^2(\theta) \]
Letting \(\theta = \sin^{-1}(x)\), we have:
\[ 1-x = 1 - 2x^2 \]

### Step 8: Simplify the equation
This simplifies to:
\[ x = 2x^2 \]
Rearranging gives:
\[ 2x^2 - x = 0 \]
Factoring out \(x\):
\[ x(2x - 1) = 0 \]

### Step 9: Solve for \(x\)
Setting each factor to zero gives:
1. \(x = 0\)
2. \(2x - 1 = 0 \Rightarrow x = \frac{1}{2}\)

### Step 10: Check the solutions
Now we need to check if both solutions are valid in the original equation:

1. For \(x = 0\):
\[ \sin^{-1}(0) + \cos^{-1}(1) = \sin^{-1}(0) \Rightarrow 0 + 0 = 0 \quad \text{(True)} \]

2. For \(x = \frac{1}{2}\):
\[ \sin^{-1}\left(\frac{1}{2}\right) + \cos^{-1}\left(\frac{1}{2}\right) = \sin^{-1}\left(-\frac{1}{2}\right) \]
\[ \frac{\pi}{6} + \frac{\pi}{3} = -\frac{\pi}{6} \Rightarrow \frac{\pi}{2} \neq -\frac{\pi}{6} \quad \text{(False)} \]

### Conclusion
The only valid solution is \(x = 0\). Therefore, the number of solutions is:
\[ \text{Number of solutions} = 1 \]