The number of solutions of sec x cos 5x + 1 = | vedclass

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(B) Given equation: $\sec x \cos 5x + 1 = 0$.Since $\sec x = \frac{1}{\cos x}$,we have $\frac{\cos 5x}{\cos x} + 1 = 0$,which implies $\cos 5x + \cos

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$8$

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(B) Given equation: $\sec x \cos 5x + 1 = 0$.Since $\sec x = \frac{1}{\cos x}$,we have $\frac{\cos 5x}{\cos x} + 1 = 0$,which implies $\cos 5x + \cos x = 0$ provided $\cos x 
eq 0$.Using the sum-to-product formula $\cos A + \cos B = 2 \cos \frac{A+B}{2} \cos \frac{A-B}{2}$,we get:$2 \cos 3x \cos 2x = 0$.This implies $\cos 3x = 0$ or $\cos 2x = 0$.For $\cos 3x = 0$ in $[0, 2\pi]$,$3x = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \frac{7\pi}{2}, \frac{9\pi}{2}, \frac{11\pi}{2}$,so $x = \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{3\pi}{2}, \frac{11\pi}{6}$.For $\cos 2x = 0$ in $[0, 2\pi]$,$2x = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \frac{7\pi}{2}$,so $x = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$.We must exclude values where $\cos x = 0$,i.e.,$x = \frac{\pi}{2}, \frac{3\pi}{2}$.The valid solutions are $x \in \{\frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}, \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}\}$.The total number of solutions is $8$.

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

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