Question

The general solution of the trigonometric equation $\tan x + \tan 2x + \tan 3x = \tan x.\tan 2x.\tan 3x$ is ?

Answer 1

Hint: We have given the question in which we have to solve and find the general solution of the trigonometric equation by somewhere using equivalent equations . Equivalent equations are said to be algebraic equations that may have the same solutions if we add or subtract the same number to both sides of an equation - Left hand side or Right hand side of the equal to sign . Or we can multiply or divide the same number to both sides of an equation - Left hand side or Right hand side of the equal to sign with the method of simplification . Also by applying the trigonometric formula .

Complete step by step answer:
In order to solve and simplify the equation we will take common \[tan{\text{ }}3x\], so we will subtract \[tan{\text{ }}3x\]both the sides of the equation .
\[
 \Rightarrow tan{\text{ }}x + tan{\text{ 2}}x + tan{\text{ }}3x = tan{\text{ }}x.tan{\text{ 2}}x.tan{\text{ }}3x \\
  \Rightarrow tan{\text{ }}x + tan{\text{ 2}}x = tan{\text{ }}x.tan{\text{ 2}}x.tan{\text{ }}3x - \tan 3x \\
   \Rightarrow tan{\text{ }}x + tan{\text{ 2}}x = \tan 3x(tan{\text{ }}x.tan{\text{ 2}}x - 1) \\
 \Rightarrow \dfrac{{tan{\text{ }}x + tan{\text{ 2}}x}}{{tan{\text{ }}x.tan{\text{ 2}}x - 1}} = \tan 3x \\
 \]
We will now take minus common from the denominator and shift it to the R.H.S.
\[\dfrac{{\tan x + \tan 2x}}{{1 - \tan x.\tan 2x}} = - \tan 3x\]
Now , we can see the resembling formula of tangent that is \[tan(A + B) = \dfrac{{\tan A + \tan B}}{{1 - \tan A.\tan B}}\]
\[tan(x + 2x) = \dfrac{{\tan x + \tan 2x}}{{1 - \tan x.\tan 2x}}\]
So, \[tan(x + 2x)\]= \[ - \tan 3x\]
\[
  tan(x + 2x) + \tan 3x = 0 \\
 \Rightarrow 2\tan 3x = 0 \\
 \Rightarrow \tan 3x = 0 \\
 \Rightarrow 3x = n\pi ,n \in \mathbb{Z} \\
 \Rightarrow x = \dfrac{{n\pi }}{3},n \in \mathbb{Z} \\
 \]
The general solution is \[x = \dfrac{{n\pi }}{3}\],\[n \in \mathbb{Z}\] .

Note: Trigonometry is one of the significant branches throughout the entire existence of mathematics and this idea is given by a Greek mathematician Hipparchus .
In equivalent equations which have identical solutions we can perform multiplication or division by the same non-zero number both L.H.S. and R.H.S. of an equation .
One must be careful while taking values from the trigonometric table and cross-check at least once to avoid any error in the answer .
Remember the trigonometric formulae and always keep the final answer simplified .