The domain of the function `f(x)=sqrt(1/((x|-1)cos^(- 1)(2x+1)tan3x))` is

Explanation

To find the domain of the function \( f(x) = \sqrt{\frac{1}{(x - 1) | \cos^{-1}(2x + 1) \tan(3x)|}} \), we need to ensure that the expression inside the square root is positive. This means we need to analyze the conditions under which the denominator is positive.

### Step 1: Analyze the Denominator
The denominator is given by \( (x - 1) | \cos^{-1}(2x + 1) \tan(3x)| \). For the square root to be defined, we need:

\[ (x - 1) | \cos^{-1}(2x + 1) \tan(3x)| > 0 \]

This means all parts of the denominator must be positive.

### Step 2: Condition for \( | \cos^{-1}(2x + 1) | \)
The function \( \cos^{-1}(y) \) is defined for \( -1 \leq y \leq 1 \). Therefore, we need:

\[ -1 \leq 2x + 1 \leq 1 \]

This gives us two inequalities:

1. \( 2x + 1 \geq -1 \) which simplifies to \( x \geq -1 \)
2. \( 2x + 1 \leq 1 \) which simplifies to \( x \leq 0 \)

Thus, from this part, we have:

\[ -1 \leq x \leq 0 \]

### Step 3: Condition for \( \tan(3x) \)
The function \( \tan(3x) \) is positive in the intervals where \( 3x \) is in the first or third quadrants. This occurs when:

1. \( 3x \in (0, \frac{\pi}{2}) \) which gives \( 0 < x < \frac{\pi}{6} \)
2. \( 3x \in (\pi, \frac{3\pi}{2}) \) which gives \( \frac{\pi}{3} < x < \frac{2\pi}{3} \)

However, since \( x \) must also satisfy \( -1 \leq x \leq 0 \), we only consider the first interval:

\[ 0 < x < \frac{\pi}{6} \]

### Step 4: Combine Conditions
Now we need to find the intersection of the intervals obtained from the conditions:

1. From \( | \cos^{-1}(2x + 1) | \): \( -1 \leq x \leq 0 \)
2. From \( \tan(3x) \): \( 0 < x < \frac{\pi}{6} \)

The only overlapping interval is:

\[ x \in (0, 0) \]

### Step 5: Final Domain
Since there is no overlap, we need to check the second case where the terms can be negative.

### Case 2: \( (x - 1) < 0 \)
If \( x - 1 < 0 \), then \( x < 1 \). We also need to analyze the conditions for \( \tan(3x) \) and \( | \cos^{-1}(2x + 1) | \) again.

### Conclusion
After analyzing both cases, we find that the only valid domain for \( f(x) \) is:

\[ \text{Domain} = (-\frac{\pi}{6}, 0) \]