The number of solutions that the equation sin | vedclass

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Question

$\sin 5 	heta \cos 3 	heta=\sin 9 	heta \cos 7 	heta$

$\Rightarrow 2 \sin 5 	heta \cos 3 	heta=2 \sin 9 	heta \cos 7 	heta$

$\Rightarrow

Vedclass · Accepted Answer

$5$

Vedclass · Answer

$\sin 5 	heta \cos 3 	heta=\sin 9 	heta \cos 7 	heta$

$\Rightarrow 2 \sin 5 	heta \cos 3 	heta=2 \sin 9 	heta \cos 7 	heta$

$\Rightarrow \sin 8 	heta+\sin 2 	heta=\sin (16 	heta)+\sin 2 	heta$

$\Rightarrow \sin 8 	heta-\sin 16 	heta=0$

$\Rightarrow \sin 8 	heta(1-2 \cos 8 	heta)=0$

$\Rightarrow \sin 8 	heta=0$ or $\cos 8 	heta=\frac{1}{2}$

$\Rightarrow 8 	heta=n \pi$ or $8 	heta=2 n \pi \pm \frac{\pi}{3}$

$\Rightarrow 	heta=\frac{n \pi}{8}$ or $	heta=\frac{2 n \pi}{8} \pm \frac{\pi}{24}$

Therefore number of values of $	heta$ is 5

The number of solutions that the equation $sin5\theta cos3\theta = sin9\theta cos7\theta $ has in $\left[ {0,\frac{\pi }{4}} \right]$ is

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