Question

How do you solve $3{\tan ^2}x - 1 = 0$?

Answer 1

In this question we have to solve the equation for $x$ , first take all constant terms to one side, take out the square root, then by using trigonometric ratios we will get the equation in the form of $\tan x = \tan \theta$ , we know that the general solution of $\tan x = \tan \theta$ is $x = n\pi + \theta$ , where $n \in Z$ $$ (i.e., $n = 0$ , $\pm 1$ , $\pm 2$ , $\pm 3$ ,…….) and $\theta \in \left( {\dfrac{{ - \pi }}{2},\dfrac{\pi }{2}} \right)$ now by substituting the values we will get the required result.

Complete step by step solution:
Given equation is $3{\tan ^2}x - 1 = 0$,
First add 1 to both sides of the equation we get,
$\Rightarrow 3{\tan ^2}x - 1 + 1 = 0 + 1$,
Now simplifying we get,
$\Rightarrow 3{\tan ^2}x = 1$,
Now first divide both sides with 3 we get,
$\Rightarrow \dfrac{{3{{\tan }^2}x}}{3} = \dfrac{1}{3}$ ,
Now simplifying we get,
$\Rightarrow {\tan ^2}x = \dfrac{1}{3}$ ,
Now taking out the square root we get,
$\Rightarrow \tan x = \pm \sqrt {\dfrac{1}{3}}$ ,
Again simplifying we get,
$\Rightarrow \tan x = \pm \dfrac{1}{{\sqrt 3 }}$ ,
Now the values for $\tan x$ are $\dfrac{1}{{\sqrt 3 }}$ and $- \dfrac{1}{{\sqrt 3 }}$ ,
Take first value i.e..,
$\Rightarrow \tan x = \dfrac{1}{{\sqrt 3 }}$ ,
Now using trigonometric ratio table we get, $\tan \dfrac{\pi }{6} = \dfrac{1}{{\sqrt 3 }}$ , so write $\dfrac{1}{{\sqrt 3 }}$ as $\tan \dfrac{\pi }{6}$ , we get,
$\Rightarrow \tan x = \tan \dfrac{\pi }{6}$ ,
So, now using the fact that the general solution of $\tan x = \tan \theta$ is $x = n\pi + \theta$ , where $n \in Z$ $$ (i.e., $n = 0$ , $\pm 1$ , $\pm 2$ , $\pm 3$ ,…….) and $\theta \in \left( {\dfrac{{ - \pi }}{2},\dfrac{\pi }{2}} \right)$ ,
Now, by substituting the values, here $\theta = \dfrac{\pi }{6}$ , in the general solution we get,
$\Rightarrow x = n\pi + \dfrac{\pi }{6}$ ,
Take second value i.e..,
$\Rightarrow \tan x = - \dfrac{1}{{\sqrt 3 }}$ ,
Now using trigonometric ratio table we get, $\tan \left( { - \dfrac{\pi }{6}} \right) = - \dfrac{1}{{\sqrt 3 }}$, so write $- \dfrac{1}{{\sqrt 3 }}$ as $\tan \left( { - \dfrac{\pi }{6}} \right)$ , we get,
$\Rightarrow \tan x = \tan \left( { - \dfrac{\pi }{6}} \right)$ ,
Since tan will have negative value in second and fourth quadrant, we can write the angle as,
$\Rightarrow \tan x = \tan \left( {\pi - \dfrac{\pi }{6}} \right)$ ,
Now simplifying we get,
$\Rightarrow \tan x = \tan \left( {\dfrac{{5\pi }}{6}} \right)$ ,
So, now using the fact that the general solution of $\tan x = \tan \theta$ is $x = n\pi + \theta$ , where $n \in Z$ $$ (i.e., $n = 0$ , $\pm 1$ , $\pm 2$ , $\pm 3$ ,…….) and $\theta \in \left( {\dfrac{{ - \pi }}{2},\dfrac{\pi }{2}} \right)$ ,
Now, by substituting the values, here $\theta = \dfrac{{5\pi }}{6}$ , in the general solution we get,
$\Rightarrow x = n\pi + \dfrac{{5\pi }}{6}$ ,
So, the general solutions are $x = n\pi + \dfrac{\pi }{6}$ and $x = n\pi + \dfrac{{5\pi }}{6}$ .

$\therefore$ The general solution for the given function i.e., $3{\tan ^2}x - 1 = 0$will be equal to $x = n\pi + \dfrac{\pi }{6}$ and $x = n\pi + \dfrac{{5\pi }}{6}$ .

Note: Trigonometric equations are those equations that involve the trigonometric functions as a variable. Principal solutions are those solutions that lie in the interval $\left[ {0,2\pi } \right]$ of such trigonometric equations, and trigonometric equation will also have general solution expressing all the values which would satisfy the given equation and it is expressed in a generalised form in terms of ‘n’.