If secx cos5x + 1 = 0,

The given equation is secx cos5x + 1 = 0.

Now,

$\sec x \cos 5 x+1=0$

$\Rightarrow \frac{\cos 5 x}{\cos x}+1=0$

$\Rightarrow \cos 5 x+\cos x=0$

$\Rightarrow 2 \cos 3 x \cos 2 x=0$

$\Rightarrow \cos 3 x=0$ or $\cos 2 x=0$

$\Rightarrow 3 x=(2 n+1) \frac{\pi}{2}, n \in \mathbf{Z}$ or $2 x=(2 m+1) \frac{\pi}{2}, m \in \mathbf{Z}$

$\Rightarrow x=(2 n+1) \frac{\pi}{6}$ or $x=(2 m+1) \frac{\pi}{4}$

Putting n = 0 and n = 1, we get

$x=\frac{\pi}{6}, \frac{\pi}{2} \quad\left(0

Also, putting m = 0, we get

$x=\frac{\pi}{4} \quad\left(0

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