If S = {x in [0, 2pi] : begin{vmatrix} 0 & co | vedclass

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(D) The given determinant is:$D = 0(0 - \sin^2 x) - \cos x(0 - \cos^2 x) - \sin x(\sin^2 x - 0) = 0$$\Rightarrow \cos^3 x - \sin^3 x = 0$$\Rightarrow

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$-4 - 2\sqrt{3}$

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(D) The given determinant is:$D = 0(0 - \sin^2 x) - \cos x(0 - \cos^2 x) - \sin x(\sin^2 x - 0) = 0$$\Rightarrow \cos^3 x - \sin^3 x = 0$$\Rightarrow \cos^3 x = \sin^3 x$$\Rightarrow 	an^3 x = 1$Since $x \in [0, 2\pi]$,the solutions for $	an x = 1$ are $x = \frac{\pi}{4}$ and $x = \frac{5\pi}{4}$.Thus,$S = \{ \frac{\pi}{4}, \frac{5\pi}{4} \}$.We need to calculate $\sum_{x \in S} 	an \left( \frac{\pi}{3} + x ight) = 	an \left( \frac{\pi}{3} + \frac{\pi}{4} ight) + 	an \left( \frac{\pi}{3} + \frac{5\pi}{4} ight)$.Using $	an(A+B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$:For $x = \frac{\pi}{4}$,$	an \left( \frac{\pi}{3} + \frac{\pi}{4} ight) = \frac{\sqrt{3} + 1}{1 - \sqrt{3}} = \frac{(\sqrt{3} + 1)^2}{1 - 3} = \frac{3 + 1 + 2\sqrt{3}}{-2} = -2 - \sqrt{3}$.For $x = \frac{5\pi}{4}$,$	an \left( \frac{\pi}{3} + \frac{5\pi}{4} ight) = 	an \left( \frac{\pi}{3} + \frac{\pi}{4} + \pi ight) = 	an \left( \frac{\pi}{3} + \frac{\pi}{4} ight) = -2 - \sqrt{3}$.Sum $= (-2 - \sqrt{3}) + (-2 - \sqrt{3}) = -4 - 2\sqrt{3}$.

If $S = \{x \in [0, 2\pi] : \begin{vmatrix} 0 & \cos x & -\sin x \\ \sin x & 0 & \cos x \\ \cos x & \sin x & 0 \end{vmatrix} = 0\}$,then $\sum_{x \in S} \tan \left( \frac{\pi}{3} + x \right)$ is equal to

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