Solve `2sin^(-1)x + sin^(-1) (1-x)= (pi)/(2)`

Answer

To solve the equation \( 2\sin^{-1}x + \sin^{-1}(1-x) = \frac{\pi}{2} \), we will follow these steps:

### Step 1: Rearranging the Equation
We start with the original equation:
\[ 2\sin^{-1}x + \sin^{-1}(1-x) = \frac{\pi}{2} \]
We can isolate \(\sin^{-1}(1-x)\) by moving it to the right side:
\[ 2\sin^{-1}x = \frac{\pi}{2} - \sin^{-1}(1-x) \]

### Step 2: Using the Inverse Trigonometric Identity
We know from the properties of inverse trigonometric functions that:
\[ \sin^{-1}y + \cos^{-1}y = \frac{\pi}{2} \]
Thus, we can replace \(\sin^{-1}(1-x)\) with \(\cos^{-1}(x)\):
\[ 2\sin^{-1}x = \cos^{-1}(1-x) \]

### Step 3: Converting to Cosine
Using the identity \(\cos^{-1}y = \frac{\pi}{2} - \sin^{-1}y\), we can rewrite the equation:
\[ 2\sin^{-1}x = \frac{\pi}{2} - \sin^{-1}(1-x) \]
This implies:
\[ \cos(2\sin^{-1}x) = 1 - x \]

### Step 4: Applying the Double Angle Formula
Using the double angle formula for cosine:
\[ \cos(2a) = 1 - 2\sin^2(a) \]
We can substitute \(a = \sin^{-1}x\):
\[ \cos(2\sin^{-1}x) = 1 - 2\sin^2(\sin^{-1}x) = 1 - 2x^2 \]
Thus, we have:
\[ 1 - 2x^2 = 1 - x \]

### Step 5: Simplifying the Equation
Now, we can simplify the equation:
\[ 1 - 2x^2 = 1 - x \]
Subtracting 1 from both sides gives:
\[ -2x^2 = -x \]
Multiplying through by -1 results in:
\[ 2x^2 - x = 0 \]

### Step 6: Factoring the Equation
Factoring out \(x\):
\[ x(2x - 1) = 0 \]
This gives us two solutions:
\[ x = 0 \quad \text{or} \quad 2x - 1 = 0 \implies x = \frac{1}{2} \]

### Step 7: Final Solutions
Thus, the solutions to the equation are:
\[ x = 0 \quad \text{and} \quad x = \frac{1}{2} \]