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

A
$(2 n + 1) \frac{π}{2}$
B
$(2 n + 1) \frac{\sqrt{π}}{5}$
C
$(2 n + 1) \frac{π}{3}$
D
$(2 n - 1) \frac{π}{10}$

Solution:

$\frac{1}{\cos 4 θ} - \frac{1}{\cos 2 θ} = 2$

$\begin{array}{ll} \Rightarrow \cos 2 θ - \cos 4 θ = 2 \cos 4 θ \cos 2 θ = \cos 6 θ + \cos 2 θ \\ ∴ \cos 6 θ + \cos 4 θ = 0 \\ \Rightarrow 2 \cos 5 θ \cos θ = 0 \\ \cos θ = 0 \Rightarrow θ = (2 n + 1) \frac{π}{2}, \cos 5 θ = 0 \Rightarrow θ = (2 n + 1) \frac{π}{10} \end{array}$

The general solution of the equation $\sec 4 θ - \sec 2 θ = 2 is θ =$

Solution:

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