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Q.
Let $A=\{x \in, R /|\sqrt{3} \cos x-\sin x| \geq 2,0 \leq x \leq 2 \pi\}$ If $x_{1} \in A, x_{2} \in A$, then $\frac{x_{1}}{x_{2}}=$
TS EAMCET 2018
Solution:
We have,
$|\sqrt{3} \cos x-\sin x| \geq 2$
It is possible only when
$\sqrt{3} \cos x-\sin x=2$ or $\sqrt{3} \cos x-\sin x=-2$
$\frac{\sqrt{3}}{2} \cos x-\frac{1}{2} \sin x=1$
or $\frac{\sqrt{3} \cos x}{2}-\frac{\sin x}{2}=-1$
$\cos \left(x+\frac{\pi}{6}\right)=1$ or $\cos \left(x+\frac{\pi}{6}\right)=-1$
$x+\frac{\pi}{6}=2 \pi$ or $x+\frac{\pi}{6}=\pi$
$\Rightarrow x=\frac{11 \pi}{6}$ or $x=\frac{5 \pi}{6}$
$\therefore $ Let $x_{1}=\frac{5 \pi}{6}$ and $x_{2}=\frac{11 \pi}{6}$
$\Rightarrow \frac{x_{1}}{x_{2}}=\frac{5}{11}$