The number of solutions of the equation 4 cos | vedclass

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Question

(A) Given equation: $4 \cos 2 	heta \cos 3 	heta = \sec 	heta$ Multiply by $\cos 	heta$ (assuming $\cos 	heta 
eq 0$): $4 \cos 2 	heta \cos 3 \

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(A) Given equation: $4 \cos 2 	heta \cos 3 	heta = \sec 	heta$ Multiply by $\cos 	heta$ (assuming $\cos 	heta 
eq 0$): $4 \cos 2 	heta \cos 3 	heta \cos 	heta = 1$ Using $2 \cos A \cos B = \cos(A+B) + \cos(A-B)$: $2 \cos 2 	heta (\cos 4 	heta + \cos 2 	heta) = 1$ $2 \cos 2 	heta \cos 4 	heta + 2 \cos^2 2 	heta = 1$ Using $2 \cos A \cos B = \cos(A+B) + \cos(A-B)$ and $2 \cos^2 A = 1 + \cos 2A$: $(\cos 6 	heta + \cos 2 	heta) + (1 + \cos 4 	heta) = 1$ $\cos 6 	heta + \cos 4 	heta + \cos 2 	heta = 0$ Using $\cos 6 	heta + \cos 2 	heta = 2 \cos 4 	heta \cos 2 	heta$: $\cos 4 	heta (2 \cos 2 	heta + 1) = 0$ Case $1$: $\cos 4 	heta = 0 \implies 4 	heta = (2n+1) \frac{\pi}{2} \implies 	heta = \frac{(2n+1) \pi}{8}$ for $n = 0, 1, \dots, 7$ ($8$ solutions). Case $2$: $\cos 2 	heta = -\frac{1}{2} \implies 2 	heta = 2n \pi \pm \frac{2 \pi}{3} \implies 	heta = n \pi \pm \frac{\pi}{3}$ for $n = 0, 1, 2$ ($4$ solutions). Total solutions = $8 + 4 = 12$.

The number of solutions of the equation $4 \cos 2 \theta \cos 3 \theta = \sec \theta$ in the interval $[0, 2 \pi]$ is

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