The general solution of the equation sec⁡4θ−sec⁡2θ=2 is θ=

# The general solution of the equation

1. A

$\left(2n+1\right)\frac{\pi }{2}$

2. B

$\left(2n+1\right)\frac{\sqrt{\pi }}{5}$

3. C

$\left(2n+1\right)\frac{\pi }{3}$

4. D

$\left(2n-1\right)\frac{\pi }{10}$

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### Solution:

$\frac{1}{\mathrm{cos}4\theta }-\frac{1}{\mathrm{cos}2\theta }=2$

$\begin{array}{l}⇒ \mathrm{cos}2\theta -\mathrm{cos}4\theta =2\mathrm{cos}4\theta \mathrm{cos}2\theta =\mathrm{cos}6\theta +\mathrm{cos}2\theta \\ \therefore \mathrm{cos}6\theta +\mathrm{cos}4\theta =0\\ ⇒ 2\mathrm{cos}5\theta \mathrm{cos}\theta =0\\ \mathrm{cos}\theta =0⇒\theta =\left(2n+1\right)\frac{\pi }{2},\mathrm{cos}5\theta =0⇒\theta =\left(2n+1\right)\frac{\pi }{10}\end{array}$

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