The domain of definition of the function `f(x)=sqrt(sin^(-1)(2x)+pi/6)` for real-valued `x` is (a)`[-1/4,1/2]` (b) `[-1/2,1/2]` (c) `(-1/2,1/9)` (d) `[-1/4,1/4]`

Explanation

To find the domain of the function \( f(x) = \sqrt{\sin^{-1}(2x) + \frac{\pi}{6}} \), we need to ensure that the expression inside the square root is non-negative. This leads us to the following steps:

### Step 1: Set up the inequality
We need the expression inside the square root to be greater than or equal to zero:
\[ \sin^{-1}(2x) + \frac{\pi}{6} \geq 0 \]

### Step 2: Isolate \(\sin^{-1}(2x)\)
Rearranging the inequality gives:
\[ \sin^{-1}(2x) \geq -\frac{\pi}{6} \]

### Step 3: Use the properties of the \(\sin^{-1}\) function
The range of the \(\sin^{-1}\) function is \([- \frac{\pi}{2}, \frac{\pi}{2}]\). Thus, we need to ensure that \(2x\) falls within the valid range of the \(\sin^{-1}\) function. Specifically, we have:
\[ -1 \leq 2x \leq 1 \]

### Step 4: Solve for \(x\)
From the inequality \(2x \geq -\frac{\pi}{6}\) and \(2x \leq 1\), we will solve these inequalities separately.

1. For \(2x \geq -\frac{\pi}{6}\):
\[ x \geq -\frac{\pi}{12} \]

2. For \(2x \leq 1\):
\[ x \leq \frac{1}{2} \]

### Step 5: Combine the inequalities
Now we combine the inequalities:
\[ -\frac{\pi}{12} \leq x \leq \frac{1}{2} \]

### Step 6: Check the range of \(2x\)
Since \(2x\) must also satisfy the range of \(\sin^{-1}\), we have:
\[ -1 \leq 2x \leq 1 \]
This translates to:
\[ -\frac{1}{2} \leq x \leq \frac{1}{2} \]

### Step 7: Find the intersection of the intervals
Now we need to find the intersection of the intervals:
1. From \(-\frac{\pi}{12} \leq x \leq \frac{1}{2}\)
2. From \(-\frac{1}{2} \leq x \leq \frac{1}{2}\)

The intersection of these intervals is:
\[ -\frac{1}{4} \leq x \leq \frac{1}{2} \]

### Conclusion
Thus, the domain of the function \( f(x) \) is:
\[ [-\frac{1}{4}, \frac{1}{2}] \]

The correct option is (a) \([-1/4, 1/2]\).