Question

If $\tan 2x = \tan \dfrac{x}{2}$, find x.

Answer 1

Hint:
We can bring all the terms of the given equation to one side. Then we can factorize the LHS using trigonometric identities. Then we can equate each of the factors to zero. Then we can solve them to obtain the required solution.

Complete step by step solution:
We are given that $\tan 2x = \tan \dfrac{x}{2}$ .
On rearranging, we get,
  $ \Rightarrow \tan 2x - \tan \dfrac{x}{2} = 0$
Now we can factorise the LHS.
We know that tangent of difference of 2 angles is given by the identity,
 $\tan \left( {A - B} \right) = \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}}$
On rearranging, we get,
 $ \Rightarrow \tan A - \tan B = \tan \left( {A - B} \right)\left( {1 + \tan A\tan B} \right)$
So, our expression will become,
 $ \Rightarrow \tan 2x - \tan \dfrac{x}{2} = \tan \left( {2x - \dfrac{x}{2}} \right)\left( {1 + \tan 2x\tan \dfrac{x}{2}} \right)$
As $\tan 2x - \tan \dfrac{x}{2} = 0$ , so we have,
 \[ \Rightarrow \tan \left( {2x - \dfrac{x}{2}} \right)\left( {1 + \tan 2x\tan \dfrac{x}{2}} \right) = 0\]
On simplification, we get,
 \[ \Rightarrow \tan \left( {\dfrac{{4x - x}}{2}} \right)\left( {1 + \tan 2x\tan \dfrac{x}{2}} \right) = 0\]
It is given that, $\tan 2x = \tan \dfrac{x}{2}$ . On applying this condition, we get,
 \[ \Rightarrow \tan \left( {\dfrac{{3x}}{2}} \right)\left( {1 + {{\tan }^2}\dfrac{x}{2}} \right) = 0\]
For the equation to satisfy, either of the factors must be equal to zero.
Either \[\tan \left( {\dfrac{{3x}}{2}} \right) = 0\] or \[\left( {1 + {{\tan }^2}\dfrac{x}{2}} \right) = 0\]
Consider the 1st case when \[\tan \left( {\dfrac{{3x}}{2}} \right) = 0\]
We know that \[\tan \left( {n\pi } \right) = 0\]
 \[ \Rightarrow \tan \left( {\dfrac{{3x}}{2}} \right) = \tan \left( {n\pi } \right)\]
On taking inverse on both sides, we get,
 \[ \Rightarrow \dfrac{{3x}}{2} = n\pi \]
On rearranging, we get,
 \[ \Rightarrow x = \dfrac{{2n\pi }}{3}\]
Now consider the 2nd case when \[\left( {1 + {{\tan }^2}\dfrac{x}{2}} \right) = 0\] .
We have \[\left( {1 + {{\tan }^2}\dfrac{x}{2}} \right) = 0\] .
On rearranging, we get,
 \[ \Rightarrow {\tan ^2}\dfrac{x}{2} = - 1\]
On taking the square root, we will get a complex value. As the trigonometric functions are real valued functions, its value cannot be a complex number. So, the equation \[\left( {1 + {{\tan }^2}\dfrac{x}{2}} \right) = 0\] has no solution and \[\left( {1 + {{\tan }^2}\dfrac{x}{2}} \right)\] cannot be zero.

So, the required solution is \[x = \dfrac{{2n\pi }}{3}\]

Note:
We must never equate the terms inside the tangent function of the given equation. For factoring the equation, we obtained, we used the trigonometric identity $\tan \left( {A - B} \right) = \dfrac{{\tan A - \tan B}}{{1 + \tan A\tan B}}$. We used the simple concept used in solving the polynomials also that an expression becomes zero when any one of the factors become zero. While solving the factors, we must substitute the given equation. Otherwise we may need to simplify it further to obtain there is no solution or an incorrect solution. We know that for real numbers and real valued functions, their squares cannot be negative. The solution we obtained is the general solution. The particular solution is given by, \[x = \dfrac{{2\pi }}{3}\].