Answer:
In the first quadrant are solutions of the form [tex]x=\dfrac{\pi}{6}+2\pi k,\ k\in Z.[/tex]
In the second quadrant are solutions of the form [tex]x=\dfrac{5\pi}{6}+2\pi k,\ k\in Z.[/tex]
Step-by-step explanation:
First, solve the equation [tex]3\tan^2x-1=0.[/tex] This equation is equivalent to equation
[tex]\tan^2 x=\dfrac{1}{3},\\ \\\tan x=\dfrac{1}{\sqrt{3} }\ \text{or }\tan x=-\dfrac{1}{\sqrt{3} }.[/tex]
The equation [tex]\tan x=\dfrac{1}{\sqrt{3} }[/tex] has the solution
[tex]x=\arctan \dfrac{1}{\sqrt{3} }+\pi k,\ k\in Z,\\ \\x=\dfrac{\pi}{6}+\pi k,\ k\in Z.[/tex]
The equation [tex]\tan x=-\dfrac{1}{\sqrt{3} }[/tex] has the solution
[tex]x=\arctan \left(-\dfrac{1}{\sqrt{3} }\right)+\pi k,\ k\in Z,\\ \\x=-\dfrac{\pi}{6}+\pi k,\ k\in Z.[/tex]
In the first quadrant are solutions of the form [tex]x=\dfrac{\pi}{6}+2\pi k,\ k\in Z.[/tex]
In the second quadrant are solutions of the form [tex]x=\dfrac{5\pi}{6}+2\pi k,\ k\in Z.[/tex]