Let A = {x in R : |sqrt{3} cos x - sin x| geq | vedclass

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(C) Given the inequality $|\sqrt{3} \cos x - \sin x| \geq 2$. We know that the maximum value of $a \cos x + b \sin x$ is $\sqrt{a^2 + b^2}$. Here,$a =

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$\frac{5}{11}$

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(C) Given the inequality $|\sqrt{3} \cos x - \sin x| \geq 2$. We know that the maximum value of $a \cos x + b \sin x$ is $\sqrt{a^2 + b^2}$. Here,$a = \sqrt{3}$ and $b = -1$,so $\sqrt{(\sqrt{3})^2 + (-1)^2} = \sqrt{3 + 1} = 2$. Thus,the expression $|\sqrt{3} \cos x - \sin x|$ can only be $\geq 2$ when it is exactly $2$. This implies $2 \cos(x + \frac{\pi}{6}) = 2$ or $2 \cos(x + \frac{\pi}{6}) = -2$. $\cos(x + \frac{\pi}{6}) = 1 \implies x + \frac{\pi}{6} = 0, 2\pi \implies x = \frac{11\pi}{6}$ (within $[0, 2\pi]$). $\cos(x + \frac{\pi}{6}) = -1 \implies x + \frac{\pi}{6} = \pi \implies x = \frac{5\pi}{6}$. So,$A = \{\frac{5\pi}{6}, \frac{11\pi}{6}\}$. Taking $x_1 = \frac{5\pi}{6}$ and $x_2 = \frac{11\pi}{6}$,we get $\frac{x_1}{x_2} = \frac{5\pi/6}{11\pi/6} = \frac{5}{11}$.

Let $A = \{x \in R : |\sqrt{3} \cos x - \sin x| \geq 2, 0 \leq x \leq 2\pi\}$. If $x_1 \in A$ and $x_2 \in A$,then find the possible value of $\frac{x_1}{x_2}$.

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