Solve the equation for x: `sin^(-1) x+sin^(-1) (1-x)=cos^(-1) x, x ne 0`

Answer

To solve the equation \( \sin^{-1} x + \sin^{-1} (1-x) = \cos^{-1} x \) where \( x \neq 0 \), we can follow these steps:

### Step 1: Rewrite the equation
We start with the given equation:
\[ \sin^{-1} x + \sin^{-1} (1-x) = \cos^{-1} x \]

### Step 2: Use the identity for cosine inverse
Recall that:
\[ \cos^{-1} a = \frac{\pi}{2} - \sin^{-1} a \]
Applying this identity, we can rewrite the equation as:
\[ \sin^{-1} x + \sin^{-1} (1-x) = \frac{\pi}{2} - \sin^{-1} x \]

### Step 3: Rearranging the equation
Now, we can rearrange the equation by moving \( \sin^{-1} x \) to the left side:
\[ \sin^{-1} (1-x) = \frac{\pi}{2} - 2 \sin^{-1} x \]

### Step 4: Let \( \sin^{-1} x = \alpha \)
Let \( \sin^{-1} x = \alpha \). Then, we have:
\[ x = \sin \alpha \]
Substituting this into our equation gives:
\[ \sin^{-1} (1 - \sin \alpha) = \frac{\pi}{2} - 2\alpha \]

### Step 5: Use the sine identity
We know that:
\[ \sin\left(\frac{\pi}{2} - \theta\right) = \cos \theta \]
Thus, we can rewrite the left side:
\[ 1 - \sin \alpha = \sin\left(\frac{\pi}{2} - 2\alpha\right) = \cos(2\alpha) \]

### Step 6: Express \( \cos(2\alpha) \)
Using the double angle formula for cosine:
\[ \cos(2\alpha) = 1 - 2\sin^2 \alpha \]
Substituting \( \sin \alpha = x \):
\[ \cos(2\alpha) = 1 - 2x^2 \]

### Step 7: Set up the equation
Now we have:
\[ 1 - \sin \alpha = 1 - 2x^2 \]
Substituting \( \sin \alpha = x \):
\[ 1 - x = 1 - 2x^2 \]

### Step 8: Simplify the equation
Cancelling \( 1 \) from both sides gives:
\[ -x = -2x^2 \]
Rearranging this leads to:
\[ 2x^2 - x = 0 \]

### Step 9: Factor the equation
Factoring out \( x \):
\[ x(2x - 1) = 0 \]

### Step 10: Solve for \( x \)
This gives us two solutions:
1. \( x = 0 \)
2. \( 2x - 1 = 0 \) which simplifies to \( x = \frac{1}{2} \)

Since the problem states \( x \neq 0 \), we discard \( x = 0 \).

### Final Solution
Thus, the solution is:
\[ x = \frac{1}{2} \]