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

Question

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

Aicharya · Accepted Answer

Given, $\begin{vmatrix} 0 & \cos x & -\sin x \ \sin x & 0 & \cos x \ \cos x & \sin x & 0 \end{vmatrix} = 0$ Expanding the determinant, $\Rightarrow 0 \cdot (0 - \cos x \sin x) - \cos x \cdot \begin{vmatrix} 0 & -\cos^2 x \ \sin^2 x & 0 \end{vmatrix} - \sin x \cdot \begin{vmatrix} \sin^2 x & -\cos^2 x \ 0 & 0 \end{vmatrix} = 0$ $\Rightarrow \cos^3 x - \sin^3 x = 0$ $\Rightarrow 	an^3 x = 1 \Rightarrow 	an x = 1 \Rightarrow x = \frac{\pi}{4}, \frac{5\pi}{4} \Rightarrow S = \left\{ \frac{\pi}{4}, \frac{5\pi}{4} ight\}$ $	herefore \sum_{x \in S} 	an \left( \frac{\pi}{3} + x ight)$ $= \sum_{x \in S} \frac{\sqrt{3} + 	an x}{1 - \sqrt{3} 	an x}$ $= 2 \left| \frac{\sqrt{3} + 1}{1 - \sqrt{3}} ight| = -4 - 2\sqrt{3}$

If $S = \{x \in [0, 2\pi] : \begin{aligned} &0 = \cos x - \sin x, \\ &\sin x = 0, \\ &\cos x + \sin x = 0…

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