Question

If $\sec x\cos 5x+1=0$, where $0<x<2\pi$, then the value of $x$ is

• A. $\frac{\pi }{5},\frac{\pi }{4}$
• B. $\frac{\pi }{5}$
• C. $\frac{\pi }{4}$
• D. None of these

Answer 1

Answer: (C)

$\frac{\pi }{4}$

We have, $\sec x\cos 5x+1=0$
$\Rightarrow \sec x\cos 5x=-1$
$\Rightarrow \cos 5x=-\cos x$
$\Rightarrow 5x=2n\pi \pm (\pi -x)$
$\Rightarrow x=\frac{(2n+1)\pi }{6}$ or $\frac{(2n-1)\pi }{4}$
Hence, $x=\frac{\pi }{4},\frac{\pi }{2},\frac{3\pi }{4},\frac{5\pi }{6},\frac{5\pi }{4},\frac{7\pi }{6},\frac{7\pi }{4},\frac{9\pi }{6},\frac{11\pi }{6}$