If sec xcos 5x + 1 = 0, where 0

Question

(c) $\sec x\cos 5x = - 1 \Rightarrow \cos 5x = - \cos x$

$ \Rightarrow $ $5x = 2n\pi \pm (\pi - x)$

$ \Rightarrow $ $x = \frac{{(2n + 1)\pi }}{6

Vedclass · Accepted Answer

$\frac{\pi }{4}$

Vedclass · Answer

(c) $\sec x\cos 5x = - 1 \Rightarrow \cos 5x = - \cos x$

$ \Rightarrow $ $5x = 2n\pi \pm (\pi - x)$

$ \Rightarrow $ $x = \frac{{(2n + 1)\pi }}{6}{\rm{ or }}\frac{{(2n - 1)\pi }}{4}$

Hence $x = \frac{\pi }{4},\,\frac{\pi }{2},\,\frac{{3\pi }}{4},\,\frac{{5\pi }}{6},\,\frac{{5\pi }}{4},\,\frac{{7\pi }}{6},\,\frac{{7\pi }}{4},\,\frac{{9\pi }}{6},\,\frac{{11\pi }}{6}$.

If $\sec x\cos 5x + 1 = 0$, where $0 < x < 2\pi $, then $x =$

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