Question
Question: How do you solve \[3{\tan ^2}x - 1 = 0\]?...
How do you solve 3tan2x−1=0?
Solution
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 tanx=tanθ , we know that the general solution of tanx=tanθ is x=nπ+θ , where n∈Z (i.e., n=0 , ±1 , ±2 , ±3 ,…….) and θ∈(2−π,2π) now by substituting the values we will get the required result.
Complete step by step solution:
Given equation is 3tan2x−1=0,
First add 1 to both sides of the equation we get,
⇒3tan2x−1+1=0+1,
Now simplifying we get,
⇒3tan2x=1,
Now first divide both sides with 3 we get,
⇒33tan2x=31 ,
Now simplifying we get,
⇒tan2x=31 ,
Now taking out the square root we get,
⇒tanx=±31 ,
Again simplifying we get,
⇒tanx=±31 ,
Now the values for tanx are 31 and −31 ,
Take first value i.e..,
⇒tanx=31 ,
Now using trigonometric ratio table we get, tan6π=31 , so write 31 as tan6π , we get,
⇒tanx=tan6π ,
So, now using the fact that the general solution of tanx=tanθ is x=nπ+θ , where n∈Z (i.e., n=0 , ±1 , ±2 , ±3 ,…….) and θ∈(2−π,2π) ,
Now, by substituting the values, here θ=6π , in the general solution we get,
⇒x=nπ+6π ,
Take second value i.e..,
⇒tanx=−31 ,
Now using trigonometric ratio table we get, tan(−6π)=−31, so write −31 as tan(−6π) , we get,
⇒tanx=tan(−6π) ,
Since tan will have negative value in second and fourth quadrant, we can write the angle as,
⇒tanx=tan(π−6π) ,
Now simplifying we get,
⇒tanx=tan(65π) ,
So, now using the fact that the general solution of tanx=tanθ is x=nπ+θ , where n∈Z (i.e., n=0 , ±1 , ±2 , ±3 ,…….) and θ∈(2−π,2π) ,
Now, by substituting the values, here θ=65π , in the general solution we get,
⇒x=nπ+65π ,
So, the general solutions are x=nπ+6π and x=nπ+65π .
∴ The general solution for the given function i.e., 3tan2x−1=0will be equal to x=nπ+6π and x=nπ+65π .
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] 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’.