Let S={x in(-pi, pi): x neq 0, pm frac{pi}{2} | vedclass

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(C) Given equation: $\sqrt{3} \sec x+\csc x+2(	an x-\cot x)=0$$\Rightarrow \sqrt{3} \sec x+\csc x=2(\cot x-	an x)$Dividing both sides by $2$: $\frac

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(C) Given equation: $\sqrt{3} \sec x+\csc x+2(	an x-\cot x)=0$$\Rightarrow \sqrt{3} \sec x+\csc x=2(\cot x-	an x)$Dividing both sides by $2$: $\frac{\sqrt{3}}{2} \sec x +\frac{1}{2} \csc x =\cot x -	an x$Using $\sec x=\frac{1}{\cos x}, \csc x=\frac{1}{\sin x}, 	an x=\frac{\sin x}{\cos x}, \cot x=\frac{\cos x}{\sin x}$:$\frac{\sqrt{3}}{2 \cos x}+\frac{1}{2 \sin x}=\frac{\cos x}{\sin x}-\frac{\sin x}{\cos x}$Multiplying by $\sin x \cos x$:$\frac{\sqrt{3}}{2} \sin x +\frac{1}{2} \cos x =\cos^2 x -\sin^2 x$Using $\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$ and $\cos \frac{\pi}{3} = \frac{1}{2}$:$\sin \frac{\pi}{3} \sin x+\cos \frac{\pi}{3} \cos x=\cos 2 x$$\Rightarrow \cos \left(x-\frac{\pi}{3}ight)=\cos 2 x$General solution: $2x = 2n\pi \pm (x - \frac{\pi}{3})$Case $1$: $2x = 2n\pi + x - \frac{\pi}{3} \Rightarrow x = 2n\pi - \frac{\pi}{3}$For $n=0, x = -\frac{\pi}{3} \in S$.Case $2$: $2x = 2n\pi - (x - \frac{\pi}{3})$ $\Rightarrow 3x = 2n\pi + \frac{\pi}{3}$ $\Rightarrow x = \frac{2n\pi}{3} + \frac{\pi}{9}$For $n=0, x = \frac{\pi}{9} \in S$.For $n=-1, x = -\frac{2\pi}{3} + \frac{\pi}{9} = -\frac{5\pi}{9} \in S$.For $n=1, x = \frac{2\pi}{3} + \frac{\pi}{9} = \frac{7\pi}{9} \in S$.Sum of solutions: $-\frac{\pi}{3} + \frac{\pi}{9} - \frac{5\pi}{9} + \frac{7\pi}{9} = \frac{-3\pi + \pi - 5\pi + 7\pi}{9} = 0$.

Let $S=\{x \in(-\pi, \pi): x \neq 0, \pm \frac{\pi}{2}\}$. The sum of all distinct solutions of the equation $\sqrt{3} \sec x+\operatorname{cosec} x+2(\tan x-\cot x)=0$ in the set $S$ is equal to

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