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Question

Question: How do you solve \[3{\tan ^2}x - 1 = 0\]?...

How do you solve 3tan2x1=03{\tan ^2}x - 1 = 0?

Explanation

Solution

In this question we have to solve the equation for xx , 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 tanx=tanθ\tan x = \tan \theta , we know that the general solution of tanx=tanθ\tan x = \tan \theta is x=nπ+θx = n\pi + \theta , where nZn \in Z (i.e., n=0n = 0 , ±1\pm 1 , ±2\pm 2 , ±3\pm 3 ,…….) and θ(π2,π2)\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 3tan2x1=03{\tan ^2}x - 1 = 0,
First add 1 to both sides of the equation we get,
3tan2x1+1=0+1\Rightarrow 3{\tan ^2}x - 1 + 1 = 0 + 1,
Now simplifying we get,
3tan2x=1\Rightarrow 3{\tan ^2}x = 1,
Now first divide both sides with 3 we get,
3tan2x3=13\Rightarrow \dfrac{{3{{\tan }^2}x}}{3} = \dfrac{1}{3} ,
Now simplifying we get,
tan2x=13\Rightarrow {\tan ^2}x = \dfrac{1}{3} ,
Now taking out the square root we get,
tanx=±13\Rightarrow \tan x = \pm \sqrt {\dfrac{1}{3}} ,
Again simplifying we get,
tanx=±13\Rightarrow \tan x = \pm \dfrac{1}{{\sqrt 3 }} ,
Now the values for tanx\tan x are 13\dfrac{1}{{\sqrt 3 }} and 13- \dfrac{1}{{\sqrt 3 }} ,
Take first value i.e..,
tanx=13\Rightarrow \tan x = \dfrac{1}{{\sqrt 3 }} ,
Now using trigonometric ratio table we get, tanπ6=13\tan \dfrac{\pi }{6} = \dfrac{1}{{\sqrt 3 }} , so write 13\dfrac{1}{{\sqrt 3 }} as tanπ6\tan \dfrac{\pi }{6} , we get,
tanx=tanπ6\Rightarrow \tan x = \tan \dfrac{\pi }{6} ,
So, now using the fact that the general solution of tanx=tanθ\tan x = \tan \theta is x=nπ+θx = n\pi + \theta , where nZn \in Z (i.e., n=0n = 0 , ±1\pm 1 , ±2\pm 2 , ±3\pm 3 ,…….) and θ(π2,π2)\theta \in \left( {\dfrac{{ - \pi }}{2},\dfrac{\pi }{2}} \right) ,
Now, by substituting the values, here θ=π6\theta = \dfrac{\pi }{6} , in the general solution we get,
x=nπ+π6\Rightarrow x = n\pi + \dfrac{\pi }{6} ,
Take second value i.e..,
tanx=13\Rightarrow \tan x = - \dfrac{1}{{\sqrt 3 }} ,
Now using trigonometric ratio table we get, tan(π6)=13\tan \left( { - \dfrac{\pi }{6}} \right) = - \dfrac{1}{{\sqrt 3 }}, so write 13- \dfrac{1}{{\sqrt 3 }} as tan(π6)\tan \left( { - \dfrac{\pi }{6}} \right) , we get,
tanx=tan(π6)\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,
tanx=tan(ππ6)\Rightarrow \tan x = \tan \left( {\pi - \dfrac{\pi }{6}} \right) ,
Now simplifying we get,
tanx=tan(5π6)\Rightarrow \tan x = \tan \left( {\dfrac{{5\pi }}{6}} \right) ,
So, now using the fact that the general solution of tanx=tanθ\tan x = \tan \theta is x=nπ+θx = n\pi + \theta , where nZn \in Z (i.e., n=0n = 0 , ±1\pm 1 , ±2\pm 2 , ±3\pm 3 ,…….) and θ(π2,π2)\theta \in \left( {\dfrac{{ - \pi }}{2},\dfrac{\pi }{2}} \right) ,
Now, by substituting the values, here θ=5π6\theta = \dfrac{{5\pi }}{6} , in the general solution we get,
x=nπ+5π6\Rightarrow x = n\pi + \dfrac{{5\pi }}{6} ,
So, the general solutions are x=nπ+π6x = n\pi + \dfrac{\pi }{6} and x=nπ+5π6x = n\pi + \dfrac{{5\pi }}{6} .

\therefore The general solution for the given function i.e., 3tan2x1=03{\tan ^2}x - 1 = 0will be equal to x=nπ+π6x = n\pi + \dfrac{\pi }{6} and x=nπ+5π6x = 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 [0,2π]\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’.