Understanding the Problem

We are given a function f defined on (-1,1) such that f(cos4θ) = 2/(2 - sec²θ) for θ in (0, π/4) ∪ (π/4, π/2). We need to find the value(s) of f(1/3).

Let x = cos4θ. Then f(x) = 2/(2 - sec²θ). We need to find f(1/3), i.e., when x = 1/3.

Step 1: Set Up the Equation

Set cos4θ = 1/3. We need to find sec²θ (or cos²θ) in terms of cos4θ.

Step 2: Express cos4θ in Terms of cos²θ

Using the double-angle formula:

$\cos 4\theta =2{\cos }^{2}2\theta -1=2(2{\cos }^2\theta -1{)}^2-1=8{\cos }^4\theta -8{\cos }^2\theta +1$

So, $\cos 4\theta =8{\cos }^4\theta -8{\cos }^2\theta +1$

Step 3: Substitute x = cos4θ = 1/3

Let u = cos²θ. Then:

$\frac{1}{3}=8u^2-8u+1$

Multiply both sides by 3:

$1=24u^2-24u+3$

Rearrange:

$24u^2-24u+2=0$

Divide by 2:

$12u^2-12u+1=0$

Step 4: Solve the Quadratic for u

Using the quadratic formula:

$u=\frac{12\pm \sqrt{{12}^2-4\times 12\times 1}}{2\times 12}=\frac{12\pm \sqrt{144-48}}{24}=\frac{12\pm \sqrt{96}}{24}=\frac{12\pm 4\sqrt{6}}{24}=\frac{3\pm \sqrt{6}}{6}$

So, ${\cos }^2\theta =u=\frac{3\pm \sqrt{6}}{6}$

Step 5: Find sec²θ

${\sec }^2\theta =\frac{1}{{\cos }^2}=\frac{6}{3\pm \sqrt{6}}$

Step 6: Compute f(1/3) = 2/(2 - sec²θ)

Case 1: sec²θ = 6/(3 + √6)

$f(1/3)=\frac{2}{2-\frac{6}{3+\sqrt{6}}}=\frac{2}{\frac{2(3+\sqrt{6})-6}{3+\sqrt{6}}}=\frac{2(3+\sqrt{6})}{6+2\sqrt{6}-6}=\frac{2(3+\sqrt{6})}{2\sqrt{6}}=\frac{3+\sqrt{6}}{\sqrt{6}}=\frac{\sqrt{6}}{\sqrt{6}}$

But we need to rationalize:

$\frac{3}{\sqrt{6}}=\frac{3}{\sqrt{6}}\times \frac{\sqrt{6}}{\sqrt{6}}=\frac{3\sqrt{6}}{6}=\frac{\sqrt{6}}{2}$

So, $f(1/3)=1+\frac{\sqrt{6}}{2}=1+\sqrt{\frac{6}{4}}=1+\sqrt{\frac{3}{2}}$

Case 2: sec²θ = 6/(3 - √6)