Question

If $\sec x \cdot \cos 5 x+1=0$, where $0< x \leq \frac{\pi}{2}$, 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, $\sec x \cos 5 x+1=0$
$\Rightarrow \frac{1}{\cos x} \cdot \cos 5 x+\frac{1}{1}=0$
$\Rightarrow \frac{\cos 5 x+\cos x}{\cos x}=0$
$\Rightarrow \cos 5 x+\cos x=0$
$\Rightarrow 2 \cos \frac{5 x+x}{2} \cos \frac{5 x-x}{2}=0$
$\left(\because \cos C+\cos D=2 \cos \frac{C+D}{2} \cos \frac{C-D}{2}\right)$
$\Rightarrow 2 \cos 3 x \cos 2 x=0$
$\Rightarrow \cos 3 x=0$ or $\cos 2 x=0$
$\Rightarrow 3 x=(2 n+1) \frac{\pi}{2}$ or $2 x=(2 n+1) \frac{\pi}{2}$
$\Rightarrow x=(2 n+1) \frac{\pi}{6} $ or $x=(2 n+1) \frac{\pi}{4}$
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 $\cos x=0 \Rightarrow x=(2 n+1) \frac{\pi}{2} \cdot$ But $x \neq(2 n \pm 1) \frac{\pi}{2}$