If sum of all the solutions of the equation 8 | vedclass

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Question

$8 \cos x\left\{\cos \left(\frac{\pi}{6}+x\right) \cdot \cos \left(\frac{\pi}{6}-x\right)-\frac{1}{2}\right\}=1$

$\Rightarrow 4 \cos x\left\{2 \cos

Vedclass · Accepted Answer

$\frac{{13}}{9}$

Vedclass · Answer

$8 \cos x\left\{\cos \left(\frac{\pi}{6}+xight) \cdot \cos \left(\frac{\pi}{6}-xight)-\frac{1}{2}ight\}=1$

$\Rightarrow 4 \cos x\left\{2 \cos \left(\frac{\pi}{6}+xight) \cos \left(\frac{\pi}{6}-xight)-1ight\}=1$

$\Rightarrow 4 \cos x\left\{\cos 2 x+\cos \frac{\pi}{3}-1ight\}=1$

$\Rightarrow 4 \cos x\left\{\cos 2 x-\frac{1}{2}ight\}=1$

$\Rightarrow 4 \cos x\left\{2 \cos ^{2} x-1-\frac{1}{2}ight\}=1$

$\Rightarrow 8 \cos ^{3} x-6 \cos x=1$

$\Rightarrow 2\left(4 \cos ^{3} x-3 \cos xight)=1$

$ \cos 3 x=\frac{1}{2}$

$\cos 3x = \cos {60^\circ },\cos {30^\circ },\cos {420^\circ }$

$x{	ext{ }} = {20^\circ },{100^\circ },{140^\circ }$

${20^\circ } + {100^\circ } + {140^\circ }{	ext{ }} = k\pi $

$ \Rightarrow {260^\circ } = k\left( {{{180}^\circ }} ight)$

$ \Rightarrow k = \frac{{{{260}^\circ }}}{{{{180}^\circ }}} = \frac{{13}}{9}$

If sum of all the solutions of the equation $8\cos x \cdot \left( {\cos \left( {\frac{\pi }{6} + x} \right) \cdot \cos \left( {\frac{\pi }{6} - x} \right) - \frac{1}{2}} \right) = 1$ in $\left[ {0,\pi } \right]$ is $k\pi $then $k$ is equal to :

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