The principal solution of $\cot x=\sqrt{3}$ are
The principal solution of $\cot x=\sqrt{3}$ are
- $\frac{\pi}{6}, \frac{5 \pi}{6}$
- $\frac{\pi}{4}, \frac{5 \pi}{4}$
- $\frac{\pi}{6}, \frac{7 \pi}{6}$
- $\frac{\pi}{3}, \frac{7 \pi}{3}$
Solution
$\begin{aligned} & \cot x=\sqrt{3} \quad \Rightarrow \quad \tan x=\frac{1}{\sqrt{3}} \\ & \therefore \frac{1}{\sqrt{3}}=\tan \left(\pi+\frac{\pi}{6}\right)=\tan \left(\frac{\pi}{6}\right) \\ & \Rightarrow \frac{1}{\sqrt{3}}=\tan \left(\frac{7 \pi}{6}\right)=\tan \left(\frac{\pi}{6}\right)\end{aligned}$
Asked in: MHT CET 2021 (24 Sep Shift 1)
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