Question

If secx cos5x + 1 = 0, where $0<x\le \frac{\mathrm{\pi }}{2}$, find the value of x.

Answer 1

The given equation is secx cos5x + 1 = 0.
Now,
$$\begin{gathered} \sec x\cos 5x+1=0 \\ \Rightarrow \frac{\cos 5x}{\cos x}+1=0 \\ \Rightarrow \cos 5x+\cos x=0 \\ \Rightarrow 2\cos 3x\cos 2x=0 \end{gathered}$$
$$\begin{gathered} \Rightarrow \cos 3x=0\mathrm{or}\cos 2x=0 \\ \Rightarrow 3x=\left(2n+1\right)\frac{\mathrm{\pi }}{2},n\in \mathbf{Z}\mathrm{or}2x=\left(2m+1\right)\frac{\mathrm{\pi }}{2},m\in \mathbf{Z} \\ \mathbf{\Rightarrow }x=\left(2n+1\right)\frac{\mathrm{\pi }}{6}\mathrm{or}x=\left(2m+1\right)\frac{\mathrm{\pi }}{4} \end{gathered}$$
Putting n = 0 and n = 1, we get
$x=\frac{\mathrm{\pi }}{6},\frac{\mathrm{\pi }}{2}\left(0<x\le \frac{\mathrm{\pi }}{2}\right)$
Also, putting m = 0, we get
$x=\frac{\mathrm{\pi }}{4}\left(0<x\le \frac{\mathrm{\pi }}{2}\right)$

Hence, the values of x are $\frac{\mathrm{\pi }}{6},\frac{\mathrm{\pi }}{4}$ and $\frac{\mathrm{\pi }}{2}$.