Question

Solve the following equation:
\[\sec x\cos5x + 1 = 0, 0 < x < \frac{\pi}{2}\]

Answer 1

\[\sec x\cos5x + 1 = 0\]
\[ \Rightarrow \frac{\cos5x}{\cos x} + 1 = 0\]
\[ \Rightarrow \cos5x + \cos x = 0\]
\[ \Rightarrow 2\cos3x \cos2x = 0\]
\[\Rightarrow \cos3x = 0 \text{ or } \cos2x = 0\]
\[ \Rightarrow 3x = \left( 2n + 1 \right)\frac{\pi}{2}, n \in Z \text{ or }2x = \left( 2m + 1 \right)\frac{\pi}{2}, m \in Z\]
\[ \Rightarrow x = \left( 2n + 1 \right)\frac{\pi}{6} or x = \left( 2m + 1 \right)\frac{\pi}{4}\]
Putting n = 0 and m = 0, we get
\[x = \frac{\pi}{6}, \frac{\pi}{4} \left( 0 < x < \frac{\pi}{2} \right)\]