Find the range of `f(x)=sec(pi/4cos^2x),` where `-oo < x < oo

Explanation

To find the range of the function \( f(x) = \sec\left(\frac{\pi}{4} \cos^2 x\right) \) for \( -\infty < x < \infty \), we can follow these steps:

### Step 1: Determine the range of \( \cos^2 x \)
The function \( \cos^2 x \) oscillates between 0 and 1 for all real values of \( x \). Therefore, we have:
\[ 0 \leq \cos^2 x \leq 1 \]

### Step 2: Scale by \( \frac{\pi}{4} \)
Next, we multiply the entire inequality by \( \frac{\pi}{4} \):
\[ 0 \leq \frac{\pi}{4} \cos^2 x \leq \frac{\pi}{4} \]

### Step 3: Apply the secant function
Now, we apply the secant function to the entire inequality. Recall that \( \sec(\theta) = \frac{1}{\cos(\theta)} \). The secant function is defined for all angles except where the cosine is zero. Since \( \frac{\pi}{4} \cos^2 x \) will always be in the range \( [0, \frac{\pi}{4}] \), we can find the secant values:
\[ \sec(0) \leq \sec\left(\frac{\pi}{4} \cos^2 x\right) \leq \sec\left(\frac{\pi}{4}\right) \]

### Step 4: Calculate the secant values
Calculating the values:
- \( \sec(0) = 1 \)
- \( \sec\left(\frac{\pi}{4}\right) = \sec\left(45^\circ\right) = \sqrt{2} \)

Thus, we have:
\[ 1 \leq \sec\left(\frac{\pi}{4} \cos^2 x\right) \leq \sqrt{2} \]

### Step 5: Conclusion on the range
From the above inequalities, we can conclude that the range of the function \( f(x) \) is:
\[ [1, \sqrt{2}] \]

### Final Answer
The range of \( f(x) = \sec\left(\frac{\pi}{4} \cos^2 x\right) \) is \( [1, \sqrt{2}] \).

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