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Question

Question: Solve the following equation: \[\tan 3x + \tan x = 2\tan 2x\]...

Solve the following equation: tan3x+tanx=2tan2x\tan 3x + \tan x = 2\tan 2x

Explanation

Solution

In the given question, we have been given an expression involving the use of trigonometric functions. The angles are not the ones given in the range of the standard table; they are variables. We are going to solve it by converting the trigonometric functions into their primitive form. Then we are going to convert the trigonometric functions into equal forms using the appropriate formulae and then solve to get the answer.
Formula Used:
We are going to use the formula of difference of angle of tangent:
tan(AB)=tanAtanB1+tanAtanB\tan \left( {A - B} \right) = \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}}

Complete step-by-step solution:
We have to solve the following equation,
tan3x+tanx=2tan2x\tan 3x + \tan x = 2\tan 2x
Now, we can write this equation as
tan3x+tanx=tan2x+tan2x\tan 3x + \tan x = \tan 2x + \tan 2x
Rearranging the terms,
tan3xtan2x=tan2xtanx\tan 3x - \tan 2x = \tan 2x - \tan x
Multiplying and dividing by 1+tan3xtan2x1 + \tan 3x\tan 2x and 1+tan2xtanx1 + \tan 2x\tan x on the two sides,
(tan3xtan2x)(1+tan3xtan2x)(1+tan3xtan2x)=(tan2xtanx)(1+tan2xtanx)(1+tan2xtanx)\dfrac{{\left( {\tan 3x - \tan 2x} \right)\left( {1 + \tan 3x\tan 2x} \right)}}{{\left( {1 + \tan 3x\tan 2x} \right)}} = \dfrac{{\left( {\tan 2x - \tan x} \right)\left( {1 + \tan 2x\tan x} \right)}}{{\left( {1 + \tan 2x\tan x} \right)}}
Using the formula of difference of angle of tangent:
tan(AB)=tanAtanB1+tanAtanB\tan \left( {A - B} \right) = \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}}
tan(3x2x)(1+tan3xtan2x)=tan(2xx)(1+tan2xtanx)\tan \left( {3x - 2x} \right)\left( {1 + \tan 3x\tan 2x} \right) = \tan \left( {2x - x} \right)\left( {1 + \tan 2x\tan x} \right)
Solving and rearranging the terms,
tanx(1+tan3xtan2x1tan2xtanx)=0\tan x\left( {1 + \tan 3x\tan 2x - 1 - \tan 2x\tan x} \right) = 0
So, we have,
tanxtan2x(tan3xtanx)=0\tan x\tan 2x\left( {\tan 3x - \tan x} \right) = 0
Now, we get,
tanx=0\tan x = 0, tan2x=0\tan 2x = 0 and tanx=tan3x\tan x = \tan 3x
Using the standard principal results, we can say that,
x=nπx = n\pi , x=mπ2x = \dfrac{{m\pi }}{2}, where n,mZn,m \in Z

Note: In this question, we had to find the sum of given trigonometric functions. We solved this question by converting the functions into their primitive form. Then we applied the appropriate identities, used their result to get to the point where the angles of functions were equal. We have to remember that when there is no apparent identity that we can apply, we have to think of some straight-forward answer, involving the use of the basic knowledge of the subjects’ properties.