Solve `sin^(-1) x - cos^(-1) x = cos ^(-1)(sqrt3/2)`.

Answer

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

### Step 1: Identify the value of \( \cos^{-1} \left( \frac{\sqrt{3}}{2} \right) \)
We know that \( \cos \theta = \frac{\sqrt{3}}{2} \) corresponds to \( \theta = \frac{\pi}{6} \). Therefore,
\[ \cos^{-1} \left( \frac{\sqrt{3}}{2} \right) = \frac{\pi}{6} \]

### Step 2: Rewrite the equation
Now we can rewrite the original equation using the value we found:
\[ \sin^{-1} x - \cos^{-1} x = \frac{\pi}{6} \]

### Step 3: Use the identity for \( \sin^{-1} x + \cos^{-1} x \)
We know the identity:
\[ \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} \]
Let’s denote this as Equation (2).

### Step 4: Add the two equations
Now we have:
1. \( \sin^{-1} x - \cos^{-1} x = \frac{\pi}{6} \) (Equation 1)
2. \( \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} \) (Equation 2)

Adding these two equations:
\[ (\sin^{-1} x - \cos^{-1} x) + (\sin^{-1} x + \cos^{-1} x) = \frac{\pi}{6} + \frac{\pi}{2} \]
This simplifies to:
\[ 2 \sin^{-1} x = \frac{\pi}{6} + \frac{\pi}{2} \]

### Step 5: Simplify the right-hand side
To add \( \frac{\pi}{6} \) and \( \frac{\pi}{2} \), we need a common denominator:
\[ \frac{\pi}{2} = \frac{3\pi}{6} \]
Thus,
\[ \frac{\pi}{6} + \frac{3\pi}{6} = \frac{4\pi}{6} = \frac{2\pi}{3} \]
So we have:
\[ 2 \sin^{-1} x = \frac{2\pi}{3} \]

### Step 6: Solve for \( \sin^{-1} x \)
Dividing both sides by 2 gives:
\[ \sin^{-1} x = \frac{\pi}{3} \]

### Step 7: Find \( x \)
Taking the sine of both sides:
\[ x = \sin \left( \frac{\pi}{3} \right) \]
We know that:
\[ \sin \left( \frac{\pi}{3} \right) = \frac{\sqrt{3}}{2} \]
Thus, the solution is:
\[ x = \frac{\sqrt{3}}{2} \]

### Final Answer
\[ x = \frac{\sqrt{3}}{2} \]