Question

The solution of 3tan^2x-1=0 if x is in the first quadrant

Answer 1

Answer:

In the first quadrant are solutions of the form $x=\dfrac{\pi}{6}+2\pi k,\ k\in Z.$

In the second quadrant are solutions of the form $x=\dfrac{5\pi}{6}+2\pi k,\ k\in Z.$

Step-by-step explanation:

First, solve the equation $3\tan^2x-1=0.$ This equation is equivalent to equation

$\tan^2 x=\dfrac{1}{3},\\ \\\tan x=\dfrac{1}{\sqrt{3} }\ \text{or }\tan x=-\dfrac{1}{\sqrt{3} }.$

The equation $\tan x=\dfrac{1}{\sqrt{3} }$ has the solution

$x=\arctan \dfrac{1}{\sqrt{3} }+\pi k,\ k\in Z,\\ \\x=\dfrac{\pi}{6}+\pi k,\ k\in Z.$

The equation $\tan x=-\dfrac{1}{\sqrt{3} }$ has the solution

$x=\arctan \left(-\dfrac{1}{\sqrt{3} }\right)+\pi k,\ k\in Z,\\ \\x=-\dfrac{\pi}{6}+\pi k,\ k\in Z.$

In the first quadrant are solutions of the form $x=\dfrac{\pi}{6}+2\pi k,\ k\in Z.$

In the second quadrant are solutions of the form $x=\dfrac{5\pi}{6}+2\pi k,\ k\in Z.$