If S = left{ {x in left[ {0,2pi } right]:left | vedclass

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Question

since the given determinant is equal to zera

$\Rightarrow 0(0-\cos x \sin x)-\cos x\left(0-\cos ^{2} x\right)$

$-\sin x\left(\sin ^{2} x-0\right

Vedclass · Accepted Answer

$-2 - \sqrt 3 $

Vedclass · Answer

since the given determinant is equal to zera

$\Rightarrow 0(0-\cos x \sin x)-\cos x\left(0-\cos ^{2} xight)$

$-\sin x\left(\sin ^{2} x-0ight)=0$

$\Rightarrow \cos ^{3} x-\sin ^{3} x=0$

$\Rightarrow 	an ^{3}=1 \Rightarrow 	an x=1$

$	herefore \quad \sum_{x \in s} 	an \left(\frac{\pi}{3}+xight)=\sum_{x \in s} \frac{	an \pi / 3+	an x}{1-	an \pi / 3 \cdot 	an x}$

${ = \sum\limits_{x\, \in \,s} {\frac{{\sqrt 3 \, + \,1}}{{1\, - \,\sqrt 3 }}\, = \sum\limits_{x\, \in \,s} {\frac{{\sqrt 3 \, + \,1}}{{1\, - \,\sqrt 3 }}\, 	imes \,\frac{{1 + \sqrt 3 }}{{1 + \sqrt 3 }}} \,} }$

${ \Rightarrow \sum\limits_{x \in s} {\frac{{1 + 3 + 2\sqrt 3 }}{{ - 2}}}  =  - 2 - \sqrt 3 }$

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