The number of solutions that the equation sin | vedclass

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(B) Given equation: $\sin 5	heta \cos 3	heta = \sin 9	heta \cos 7	heta$Multiply both sides by $2$: $2 \sin 5	heta \cos 3	heta = 2 \sin 9	heta \

Vedclass · Accepted Answer

$5$

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(B) Given equation: $\sin 5	heta \cos 3	heta = \sin 9	heta \cos 7	heta$Multiply both sides by $2$: $2 \sin 5	heta \cos 3	heta = 2 \sin 9	heta \cos 7	heta$Using the identity $2 \sin A \cos B = \sin(A+B) + \sin(A-B)$:$\sin(5	heta + 3	heta) + \sin(5	heta - 3	heta) = \sin(9	heta + 7	heta) + \sin(9	heta - 7	heta)$$\sin 8	heta + \sin 2	heta = \sin 16	heta + \sin 2	heta$$\sin 8	heta - \sin 16	heta = 0$$\sin 8	heta - 2 \sin 8	heta \cos 8	heta = 0$$\sin 8	heta (1 - 2 \cos 8	heta) = 0$Case $1$: $\sin 8	heta = 0 \Rightarrow 8	heta = n\pi \Rightarrow 	heta = \frac{n\pi}{8}$.For $	heta \in [0, \frac{\pi}{4}]$,$n = 0, 1, 2$. Values are $0, \frac{\pi}{8}, \frac{\pi}{4}$.Case $2$: $\cos 8	heta = \frac{1}{2} \Rightarrow 8	heta = 2n\pi \pm \frac{\pi}{3} \Rightarrow 	heta = \frac{n\pi}{4} \pm \frac{\pi}{24}$.For $	heta \in [0, \frac{\pi}{4}]$,$n=0$ gives $	heta = \frac{\pi}{24}$ and $n=1$ gives $	heta = \frac{\pi}{4} - \frac{\pi}{24} = \frac{5\pi}{24}$.The set of solutions is ${0, \frac{\pi}{24}, \frac{\pi}{8}, \frac{5\pi}{24}, \frac{\pi}{4}}$.Total number of solutions is $5$.

The number of solutions that the equation $\sin 5\theta \cos 3\theta = \sin 9\theta \cos 7\theta$ has in the interval $[0, \frac{\pi}{4}]$ is

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