Question

If \(sec4θ-sec2θ = 2\), then \(θ =\)

• A. \(nπ+\frac{π}{8},\frac{nπ}{5}+\frac{π}{6},n\in Z\)
• B. \(nπ+\frac{π}{6},\frac{nπ}{5}+\frac{π}{8},n\in Z\)
• C. \(nπ+\frac{π}{2},\frac{nπ}{5}+\frac{π}{10},n\in Z\)
• D. \(nπ+\frac{π}{3},nπ+\frac{π}{10},n\in Z\)

Hint:

Put c = cos 2theta, use cos 4theta = 2c^2 - 1 and factor the cubic.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
Convert to cosines. Let \(c = \cos 2\theta\), so \(\cos 4\theta = 2c^2 - 1\).

Step 2: Form the equation
\[ \frac{1}{\cos4\theta} - \frac{1}{\cos2\theta} = 2 \Rightarrow \cos2\theta - \cos4\theta = 2\cos4\theta\cos2\theta \]
\[ c - (2c^2 - 1) = 2c(2c^2 - 1) \Rightarrow 4c^3 + 2c^2 - 3c - 1 = 0 \]
Try \(c = -1\): \(-4 + 2 + 3 - 1 = 0\). So \((c + 1)(4c^2 - 2c - 1) = 0\).

Step 3: Solve each factor
\(c = -1\): \(2\theta = (2n+1)\pi\), so \(\theta = n\pi + \frac{\pi}{2}\).
\(4c^2 - 2c - 1 = 0\): \(c = \frac{1 \pm \sqrt5}{4}\), which is \(\cos 36^{\circ}\) or \(\cos 108^{\circ}\). Combining \(2\theta = 2n\pi \pm \frac{\pi}{5}\) and \(2\theta = 2n\pi \pm \frac{3\pi}{5}\) gives \(\theta = \frac{n\pi}{5} + \frac{\pi}{10}\).
So the solution is \(\theta = n\pi + \frac{\pi}{2}, \frac{n\pi}{5} + \frac{\pi}{10}\). This is option (C). The other options have angles such as \(\pi/8\) or \(\pi/6\) that do not satisfy the equation.

Final Answer:
The solution set is option (C). \[ \boxed{n\pi + \frac{\pi}{2},\ \frac{n\pi}{5} + \frac{\pi}{10}} \]