Find the range of `f(x)=1/(1-3sqrt(1-sin^2x))`

Explanation

To find the range of the function \( f(x) = \frac{1}{1 - 3\sqrt{1 - \sin^2 x}} \), we can follow these steps:

### Step 1: Rewrite the function
We start by rewriting \( \sqrt{1 - \sin^2 x} \) as \( \cos x \):
\[ f(x) = \frac{1}{1 - 3\sqrt{1 - \sin^2 x}} = \frac{1}{1 - 3\cos x} \]

### Step 2: Determine the range of \( \cos x \)
The cosine function \( \cos x \) oscillates between -1 and 1. Therefore, we have:
\[ -1 \leq \cos x \leq 1 \]

### Step 3: Transform the range of \( \cos x \) to \( 3\cos x \)
Multiplying the entire inequality by 3, we get:
\[ -3 \leq 3\cos x \leq 3 \]

### Step 4: Adjust the inequality for \( 1 - 3\cos x \)
Next, we adjust the inequality to find the range of \( 1 - 3\cos x \):
\[ 1 - 3 \leq 1 - 3\cos x \leq 1 + 3 \]
This simplifies to:
\[ -2 \leq 1 - 3\cos x \leq 4 \]

### Step 5: Find the reciprocal to determine the range of \( f(x) \)
Now we need to find the range of \( f(x) = \frac{1}{1 - 3\cos x} \). We will consider the values of \( 1 - 3\cos x \) in the intervals we found:
1. When \( 1 - 3\cos x = -2 \), \( f(x) = \frac{1}{-2} = -\frac{1}{2} \).
2. When \( 1 - 3\cos x = 4 \), \( f(x) = \frac{1}{4} \).

### Step 6: Analyze the behavior of \( f(x) \)
We need to analyze the behavior of \( f(x) \) as \( 1 - 3\cos x \) approaches its limits:
- As \( 1 - 3\cos x \) approaches -2 from the right, \( f(x) \) approaches -\(\frac{1}{2}\).
- As \( 1 - 3\cos x \) approaches 0 (which happens when \( \cos x = \frac{1}{3} \)), \( f(x) \) approaches \( +\infty \).
- As \( 1 - 3\cos x \) approaches 4, \( f(x) \) approaches \( \frac{1}{4} \).

### Step 7: Combine the results
Thus, the range of \( f(x) \) can be expressed as:
\[ (-\infty, -\frac{1}{2}) \cup \left(\frac{1}{4}, \infty\right) \]

### Final Answer
The range of the function \( f(x) = \frac{1}{1 - 3\sqrt{1 - \sin^2 x}} \) is:
\[ (-\infty, -\frac{1}{2}) \cup \left(\frac{1}{4}, \infty\right) \]