Question

The general solution(s) of the equation $\sec 4\theta -\sec 2\theta =2$ can be

• A. $(2n+1)\frac{\pi }{12},n\in Z$
• B. $(2n+1)\frac{\pi }{10},n\in Z$
• C. $(2n+1)\frac{\pi }{2},n\in Z$
• D. $(2n+1)\frac{\pi }{4},n\in Z$

Answer 1

Answer: (B), (C)

$(2n+1)\frac{\pi }{2},n\in Z$

$\sec 4\theta -\sec 2\theta =2$
$\Rightarrow \frac{1}{\cos 4\theta }-\frac{1}{\cos 2\theta }=2$
$\Rightarrow \cos 2\theta -\cos 4\theta =2\cos 2\theta .\cos 4\theta$
$\Rightarrow \cos 2\theta -\cos 4\theta =\cos 6\theta +\cos 2\theta$
$\Rightarrow \cos 6\theta +\cos 4\theta =0$
$\Rightarrow 2\cos 5\theta .\cos \theta =0$
$\Rightarrow \cos 5\theta =0\text{or}\cos \theta =0$
$\Rightarrow 5\theta ,\theta =(2n+1)\frac{\pi }{2}\therefore \theta \in \left\{\right(2n+1\left)\frac{\pi }{2}\right\}\cup \left\{\right(2n+1\left)\frac{\pi }{10}\right\},n\in Z$