The true set of valued of 'K' for which `sin ^(-1)((1)/(1+sin ^(2)x)) = (kpi)/(6)` may have a solution is :

Options

A. `[(1)/(4), (1)/(2)]`
B. `[1,2]`
C. `[(1)/(6), (1)/(2)]`
D. `[2,4]`

Answer

To solve the equation \( \sin^{-1}\left(\frac{1}{1+\sin^2 x}\right) = \frac{k\pi}{6} \) and find the true set of values of \( k \) for which this equation may have a solution, we can follow these steps:

### Step 1: Determine the range of \( \sin^2 x \)
The function \( \sin^2 x \) has a minimum value of 0 and a maximum value of 1. Therefore, we can express this as:
\[ 0 \leq \sin^2 x \leq 1 \]

### Step 2: Find the range of \( 1 + \sin^2 x \)
Adding 1 to \( \sin^2 x \):
\[ 1 \leq 1 + \sin^2 x \leq 2 \]

### Step 3: Determine the range of \( \frac{1}{1 + \sin^2 x} \)
Taking the reciprocal of the range \( [1, 2] \):
\[ \frac{1}{2} \leq \frac{1}{1 + \sin^2 x} \leq 1 \]

### Step 4: Find the range of \( \sin^{-1}\left(\frac{1}{1 + \sin^2 x}\right) \)
Now we apply the \( \sin^{-1} \) function to the range \( \left[\frac{1}{2}, 1\right] \):
- The minimum value occurs at \( \frac{1}{2} \):
\[ \sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6} \]
- The maximum value occurs at 1:
\[ \sin^{-1}(1) = \frac{\pi}{2} \]

Thus, the range of \( \sin^{-1}\left(\frac{1}{1 + \sin^2 x}\right) \) is:
\[ \frac{\pi}{6} \leq \sin^{-1}\left(\frac{1}{1 + \sin^2 x}\right) \leq \frac{\pi}{2} \]

### Step 5: Set the equation with the range
We have:
\[ \frac{\pi}{6} \leq \frac{k\pi}{6} \leq \frac{\pi}{2} \]

### Step 6: Solve for \( k \)
Dividing the entire inequality by \( \frac{\pi}{6} \):
\[ 1 \leq k \leq 3 \]

### Conclusion
Thus, the true set of values of \( k \) for which the equation has a solution is:
\[ k \in [1, 3] \]