Solve the following equations : `sin^(-1)(x^(2)-2x+3)+cos^(-1)(x^(2)-x)=(pi)/2`

Explanation

To solve the equation \( \sin^{-1}(x^2 - 2x + 3) + \cos^{-1}(x^2 - x) = \frac{\pi}{2} \), we can follow these steps:

### Step 1: Use the identity for inverse sine and cosine
We know that:
\[ \sin^{-1}(y) + \cos^{-1}(y) = \frac{\pi}{2} \]
for any \( y \) in the range of the sine and cosine functions.

### Step 2: Set the arguments equal
From the given equation, we can equate the arguments of the sine and cosine functions:
\[ x^2 - 2x + 3 = x^2 - x \]

### Step 3: Simplify the equation
Now, we simplify the equation:
\[ x^2 - 2x + 3 - (x^2 - x) = 0 \]
This simplifies to:
\[ -2x + 3 + x = 0 \]
which further simplifies to:
\[ -x + 3 = 0 \]

### Step 4: Solve for \( x \)
Now, we can solve for \( x \):
\[ -x = -3 \implies x = 3 \]

### Conclusion
Thus, the solution to the equation is:
\[ \boxed{3} \]