Mathematics | MHT CET 2020 | Discussion Page

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The principal solutions of $\cot x=\sqrt{3}$ are

$\frac{\pi}{6},\frac{7\pi}{6}$

$\frac{\pi}{3},\frac{7\pi}{3}$

$\frac{\pi}{4},\frac{5\pi}{4}$

$\frac{\pi}{6},\frac{5\pi}{6}$

Option A

we have,

$\cot x=\sqrt{3}$

$\cot x=\cot\frac{\pi}{6}$ or $\cot\left(\pi+\frac{\pi}{6}\right)$

$\therefore$ $x=\frac{\pi}{6} ,\frac{7\pi}{6}$

$\therefore$ principal solution is $\frac{\pi}{6},\frac{7\pi}{6}$

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