The number of solutions of the equation sec x | vedclass

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(B) Given the equation $\sec x \cos 5x + 1 = 0$,we can rewrite it as $\frac{\cos 5x}{\cos x} = -1$,provided $\cos x 
eq 0$.This implies $\cos 5x = -\

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(B) Given the equation $\sec x \cos 5x + 1 = 0$,we can rewrite it as $\frac{\cos 5x}{\cos x} = -1$,provided $\cos x 
eq 0$.This implies $\cos 5x = -\cos x$,which is equivalent to $\cos 5x = \cos(\pi - x)$.The general solution for $\cos 	heta = \cos \alpha$ is $	heta = 2n\pi \pm \alpha$.Case $1$: $5x = 2n\pi + (\pi - x) \implies 6x = (2n+1)\pi \implies x = \frac{(2n+1)\pi}{6}$.For $x \in [0, 2\pi]$,$n = 0, 1, 2, 3, 4, 5$ gives $x = \frac{\pi}{6}, \frac{3\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{9\pi}{6}, \frac{11\pi}{6}$.Checking $\cos x 
eq 0$: $\frac{3\pi}{6} = \frac{\pi}{2}$ and $\frac{9\pi}{6} = \frac{3\pi}{2}$ make $\cos x = 0$,so these are excluded.Remaining values: $\frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}$ ($4$ solutions).Case $2$: $5x = 2n\pi - (\pi - x) \implies 4x = (2n-1)\pi \implies x = \frac{(2n-1)\pi}{4}$.For $x \in [0, 2\pi]$,$n = 1, 2, 3, 4$ gives $x = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$.None of these make $\cos x = 0$ ($4$ solutions).Total solutions = $4 + 4 = 8$.

The number of solutions of the equation $\sec x \cos 5x + 1 = 0$ in the interval $[0, 2\pi]$ is

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