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: Substitute
With $c = \cos 2\theta$ the equation becomes $c - \cos 4\theta = 2 c \cos 4\theta$.

Step 2: Cubic
$4c^3 + 2c^2 - 3c - 1 = (c+1)(4c^2 - 2c - 1) = 0$.

Step 3: Back to theta
$c = -1$ gives $\theta = n\pi + \pi/2$. The quadratic gives $\cos 2\theta = \cos(\pi/5)$ or $\cos(3\pi/5)$, which together are $\theta = n\pi/5 + \pi/10$. Option (C).

Final Answer:
Option (C). \[ \boxed{n\pi + \frac{\pi}{2},\ \frac{n\pi}{5} + \frac{\pi}{10}} \]