Solve : `cos^(-1) (sin cos^(-1)x ) =(pi)/(6)`.

Answer

To solve the equation \( \cos^{-1}(\sin(\cos^{-1} x)) = \frac{\pi}{6} \), we will follow these steps:

### Step 1: Use the property of inverse trigonometric functions
We know that:
\[ \cos^{-1} y + \sin^{-1} y = \frac{\pi}{2} \]
Thus, we can rewrite \( \cos^{-1} x \) as:
\[ \cos^{-1} x = \frac{\pi}{2} - \sin^{-1} x \]
Substituting this into our equation gives:
\[ \cos^{-1}(\sin(\frac{\pi}{2} - \sin^{-1} x)) = \frac{\pi}{6} \]

### Step 2: Simplify using the sine and cosine relationship
Using the identity \( \sin(\frac{\pi}{2} - \theta) = \cos(\theta) \), we have:
\[ \sin(\frac{\pi}{2} - \sin^{-1} x) = \cos(\sin^{-1} x) \]
Thus, our equation becomes:
\[ \cos^{-1}(\cos(\sin^{-1} x)) = \frac{\pi}{6} \]

### Step 3: Cancel the inverse cosine and cosine
Since \( \cos^{-1}(\cos y) = y \) for \( y \) in the range \( [0, \pi] \), we can simplify to:
\[ \sin^{-1} x = \frac{\pi}{6} \]

### Step 4: Solve for \( x \)
Taking the sine of both sides gives:
\[ x = \sin\left(\frac{\pi}{6}\right) \]
We know that:
\[ \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \]
Thus, we find:
\[ x = \frac{1}{2} \]

### Final Answer
The solution to the equation is:
\[ x = \frac{1}{2} \]