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

Question

(D) Given $\sec x \cos 5x + 1 = 0$,we have $\cos 5x = -\cos x = \cos(\pi - x)$.General solution for $\cos 	heta = \cos \alpha$ is $	heta = 2n\pi \pm

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None of these

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(D) Given $\sec x \cos 5x + 1 = 0$,we have $\cos 5x = -\cos x = \cos(\pi - x)$.General solution for $\cos 	heta = \cos \alpha$ is $	heta = 2n\pi \pm \alpha$.Case $1$: $5x = 2n\pi + (\pi - x)$ $\Rightarrow 6x = (2n + 1)\pi$ $\Rightarrow x = \frac{(2n + 1)\pi}{6}$.For $n=0, 1, 2, 3, 4, 5$,$x = \frac{\pi}{6}, \frac{3\pi}{6} (\frac{\pi}{2}), \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{9\pi}{6} (\frac{3\pi}{2}), \frac{11\pi}{6}$.Case $2$: $5x = 2n\pi - (\pi - x)$ $\Rightarrow 4x = 2n\pi - \pi$ $\Rightarrow x = \frac{(2n - 1)\pi}{4}$.For $n=1, 2, 3, 4$,$x = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$.Since the provided options do not match the complete solution set,the correct choice is $D$.

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

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