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

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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}$

${ = \sum\limits_{x\, \in \,s} {\frac{{\sqrt 3 \, + \,1}}{{1\, - \,\sqrt 3 }}\, = \sum\limits_{x\, \in \,s} {\frac{{\sqrt 3 \, + \,1}}{{1\, - \,\sqrt 3 }}\, 	imes \,\frac{{1 + \sqrt 3 }}{{1 + \sqrt 3 }}} \,} }$

${ \Rightarrow \sum\limits_{x \in s} {\frac{{1 + 3 + 2\sqrt 3 }}{{ - 2}}}  =  - 2 - \sqrt 3 }$

Similar Questions

The equation ${\sin ^4}x + {\cos ^4}x + \sin 2x + \alpha = 0$ is solvable for

If $\cos A\sin \left( {A - \frac{\pi }{6}} \right)$ is maximum, then the value of $A$ is equal to

If $\cos 7\theta = \cos \theta - \sin 4\theta $, then the general value of $\theta $ is

The number of solutions to $\sin x=\frac{6}{x}$ with $0 \leq x \leq 12 \pi$ is

The number of solutions of the equation $\sin (9 x)+\sin (3 x)=0$ in the closed interval $[0,2 \pi]$ is