If mathrm{n} is the number of solutions of th | vedclass

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Question

$2 \cos x\left(4 \sin \left(\frac{\pi}{4}+x\right) \sin \left(\frac{\pi}{4}-x\right)-1\right)=1$

$2 \cos x\left(4\left(\sin ^{2} \frac{\pi}{4}-\sin

Vedclass · Accepted Answer

$(3,13 \pi / 3)$

Vedclass · Answer

$2 \cos x\left(4 \sin \left(\frac{\pi}{4}+xight) \sin \left(\frac{\pi}{4}-xight)-1ight)=1$

$2 \cos x\left(4\left(\sin ^{2} \frac{\pi}{4}-\sin ^{2} xight)-1ight)=1$

$2 \cos x\left(4\left(\frac{1}{2}-\sin ^{2} xight)-1ight)=1$

$2 \cos x\left(2-4 \sin ^{2} x-1ight)=1$

$2 \cos x\left(1-4 \sin ^{2} xight)=1$

$2 \cos x\left(4 \cos ^{2} x-3ight)=1$

$4 \cos ^{3} x-3 \cos x=\frac{1}{2}$

$\cos 3 x=\frac{1}{2}$

$\mathrm{x} \in[0, \pi] 	herefore 3 \mathrm{x} \in[0,3 \pi]$

If $\mathrm{n}$ is the number of solutions of the equation

$2 \cos x\left(4 \sin \left(\frac{\pi}{4}+x\right) \sin \left(\frac{\pi}{4}-x\right)-1\right)=1, x \in[0, \pi]$

and $S$ is the sum of all these solutions, then the ordered pair $(\mathrm{n}, \mathrm{S})$ is :

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If $\mathrm{n}$ is the number of solutions of the equation $2 \cos x\left(4 \sin \left(\frac{\pi}{4}+x\right) \sin \left(\frac{\pi}{4}-x\right)-1\right)=1, x \in[0, \pi]$ and $S$ is the sum of all these solutions, then the ordered pair $(\mathrm{n}, \mathrm{S})$ is :