Question

If $\sec x\cdot \cos 5x+1=0$, where $0<x\le \frac{\pi }{2}$, then the value of $x$ is/are

• A. $\frac{\pi }{6}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{2}$
• D. Both(a) and (b)

Answer 1

Answer: (D)

Given, $\sec x\cos 5x+1=0$
$\Rightarrow \frac{1}{\cos x}\cdot \cos 5x+\frac{1}{1}=0$
$\Rightarrow \frac{\cos 5x+\cos x}{\cos x}=0$
$\Rightarrow \cos 5x+\cos x=0$
$\Rightarrow 2\cos \frac{5x+x}{2}\cos \frac{5x-x}{2}=0$
$\left(\because \cos C+\cos D=2\cos \frac{C+D}{2}\cos \frac{C-D}{2}\right)$
$\Rightarrow 2\cos 3x\cos 2x=0$
$\Rightarrow \cos 3x=0$ or $\cos 2x=0$
$\Rightarrow 3x=(2n+1)\frac{\pi }{2}$ or $2x=(2n+1)\frac{\pi }{2}$
$\Rightarrow x=(2n+1)\frac{\pi }{6}$ or $x=(2n+1)\frac{\pi }{4}$
If $n=0$, then $x=\frac{\pi }{6}$ or $x=\frac{\pi }{4}$
If $n=1$, then
$x=3\times \frac{\pi }{6}=\frac{\pi }{2}$
or $x=\frac{3\pi }{4}>\frac{\pi }{2}$
Therefore, the values of $x$ are $\frac{\pi }{6},\frac{\pi }{4}$.
Note $\cos x=0\Rightarrow x=(2n+1)\frac{\pi }{2}\cdot$ But $x\ne (2n\pm 1)\frac{\pi }{2}$