If `sin^(-1)x+cos^(-1)(1-x)=sin^(-1)(-x)`, then x satisfies the equation

Options

A. `2x^(2)-x+2=0`
B. `2x^(2)-3x=0`
C. `2x^(2)+x-1=0`
D. None of these

Answer

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

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

### Step 2: Use the Identity for \(\cos^{-1}\)
Recall that:
\[ \cos^{-1} y = \frac{\pi}{2} - \sin^{-1} y \]
Using this identity, we can rewrite \(\cos^{-1} (1 - x)\):
\[ \cos^{-1} (1 - x) = \frac{\pi}{2} - \sin^{-1} (1 - x) \]

### Step 3: Substitute Back into the Equation
Substituting this into the original equation gives:
\[ \sin^{-1} x + \left( \frac{\pi}{2} - \sin^{-1} (1 - x) \right) = \sin^{-1} (-x) \]
This simplifies to:
\[ \sin^{-1} x + \frac{\pi}{2} - \sin^{-1} (1 - x) = \sin^{-1} (-x) \]

### Step 4: Rearranging the Equation
Rearranging the equation, we have:
\[ \sin^{-1} x - \sin^{-1} (1 - x) = \sin^{-1} (-x) - \frac{\pi}{2} \]

### Step 5: Use the Identity for \(\sin^{-1} (-x)\)
We know that:
\[ \sin^{-1} (-x) = -\sin^{-1} x \]
Substituting this into our equation gives:
\[ \sin^{-1} x - \sin^{-1} (1 - x) = -\sin^{-1} x - \frac{\pi}{2} \]

### Step 6: Combine Like Terms
Combining like terms results in:
\[ 2\sin^{-1} x = -\sin^{-1} (1 - x) - \frac{\pi}{2} \]

### Step 7: Solve for \(x\)
To solve for \(x\), we can express \(\sin^{-1} (1 - x)\) in terms of \(x\):
\[ \sin^{-1} (1 - x) = \frac{\pi}{2} + 2\sin^{-1} x \]
This leads us to:
\[ 1 - x = \sin\left(\frac{\pi}{2} + 2\sin^{-1} x\right) \]
Using the sine addition formula, we can simplify this further.

### Step 8: Find Values of \(x\)
After solving the equation, we find two potential solutions:
1. \(x = 0\)
2. \(x = \frac{1}{2}\)

### Step 9: Verify the Solutions
We substitute \(x = 0\) and \(x = \frac{1}{2}\) back into the original equation to check which one satisfies it:
- For \(x = 0\):
\[ \sin^{-1}(0) + \cos^{-1}(1) = \sin^{-1}(0) \quad \Rightarrow \quad 0 + 0 = 0 \quad \text{(True)} \]
- For \(x = \frac{1}{2}\):
\[ \sin^{-1}\left(\frac{1}{2}\right) + \cos^{-1}\left(\frac{1}{2}\right) = \sin^{-1}\left(-\frac{1}{2}\right) \quad \Rightarrow \quad \frac{\pi}{6} + \frac{\pi}{3} \neq -\frac{\pi}{6} \quad \text{(False)} \]

Thus, the only solution that satisfies the equation is:
\[ \boxed{0} \]