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$x=\frac{5\pi}{6}+k\pi$ $\newline$ Solve the equation in the interval 0 \leq x < 2\pi. $\newline$ $3\tan x-\sqrt{3}=0$

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Q.

x=\frac{5\pi}{6}+k\pi

\newline

Solve the equation in the interval

0 \leq x < 2\pi

.

\newline

3\tan x-\sqrt{3}=0

Question Prompt: Question Prompt: Solve the equation $3\tan x - \sqrt{3} = 0$ for $x$ in the interval 0 \leq x < 2\pi, given $x = \left(\frac{5\pi}{6}\right) + k\pi$ .
Step $1$ : Step $1$ : Simplify the given equation to find $x$ . $3\tan x = \sqrt{3}$ $\tan x = \frac{\sqrt{3}}{3}$ This implies $x = \arctan(\frac{\sqrt{3}}{3})$ .

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