Find the range of `f(x)=sqrt(4-sqrt(1+tan^2x))`.

Explanation

To find the range of the function \( f(x) = \sqrt{4 - \sqrt{1 + \tan^2 x}} \), we can follow these steps:

### Step 1: Simplify the expression inside the square root
We know that \( 1 + \tan^2 x = \sec^2 x \). Therefore, we can rewrite the function as:
\[ f(x) = \sqrt{4 - \sqrt{\sec^2 x}} \]

### Step 2: Rewrite \( \sqrt{\sec^2 x} \)
Since \( \sec x = \frac{1}{\cos x} \), we have:
\[ \sqrt{\sec^2 x} = |\sec x| \]
Thus, the function becomes:
\[ f(x) = \sqrt{4 - |\sec x|} \]

### Step 3: Determine the range of \( |\sec x| \)
The secant function \( \sec x \) has a range of \( (-\infty, -1] \cup [1, \infty) \). Therefore, the absolute value \( |\sec x| \) has a range of:
\[ [1, \infty) \]

### Step 4: Analyze the expression \( 4 - |\sec x| \)
Since \( |\sec x| \geq 1 \), we can find the minimum value of \( 4 - |\sec x| \):
\[ 4 - |\sec x| \leq 4 - 1 = 3 \]
Thus, the maximum value of \( 4 - |\sec x| \) is 3, and it can decrease without bound as \( |\sec x| \) increases.

### Step 5: Determine the range of \( f(x) \)
Now, we need to find the range of \( f(x) = \sqrt{4 - |\sec x|} \). Since \( 4 - |\sec x| \) can take values from \( 3 \) (when \( |\sec x| = 1 \)) to \( -\infty \) (as \( |\sec x| \) approaches infinity), we have:
\[ 4 - |\sec x| \in (-\infty, 3] \]
However, since we are taking the square root, we need to ensure that \( 4 - |\sec x| \) is non-negative:
\[ 4 - |\sec x| \geq 0 \implies |\sec x| \leq 4 \]
This means \( |\sec x| \) can take values from \( 1 \) to \( 4 \).

### Step 6: Find the corresponding values of \( f(x) \)
When \( |\sec x| = 1 \):
\[ f(x) = \sqrt{4 - 1} = \sqrt{3} \]
When \( |\sec x| = 4 \):
\[ f(x) = \sqrt{4 - 4} = 0 \]
Thus, the function \( f(x) \) can take values from \( 0 \) to \( \sqrt{3} \).

### Conclusion
The range of the function \( f(x) \) is:
\[ [0, \sqrt{3}] \]