Question

If secx cos5x + 1 = 0, where \[0 < x \leq \frac{\pi}{2}\], find the value of x.

Answer 1

The given equation is secx cos5x + 1 = 0.
Now,
\[\sec x\cos5x + 1 = 0\]
\[ \Rightarrow \frac{\cos5x}{\cos x} + 1 = 0\]
\[ \Rightarrow \cos5x + \cos x = 0\]
\[ \Rightarrow 2\cos3x \cos2x = 0\]
\[\Rightarrow \cos3x = 0\text{ or }\cos2x = 0\]
\[ \Rightarrow 3x = \left( 2n + 1 \right)\frac{\pi}{2}, n \in Z\text{ or }2x = \left( 2m + 1 \right)\frac{\pi}{2}, m \in Z\]
\[ \Rightarrow x = \left( 2n + 1 \right)\frac{\pi}{6}\text{ or }x = \left( 2m + 1 \right)\frac{\pi}{4}\]
Putting n = 0 and n = 1, we get
\[x = \frac{\pi}{6}, \frac{\pi}{2} \left( 0 < x \leq \frac{\pi}{2} \right)\]
Also, putting m = 0, we get \[x = \frac{\pi}{4} \left( 0 < x \leq \frac{\pi}{2} \right)\]
Hence, the values of x are \[\frac{\pi}{6}, \frac{\pi}{4}\] and \[\frac{\pi}{2}\].