Question

If $\tan 3x+\tan x=2\tan 2x$ then $x$ is equal to $(n\in Z)$

• A. $n\pi$
• B. $\frac{n\pi }{4}$
• C. $2n\pi$
• D. $\frac{3n\pi }{4}$

Answer 1

The correct option is A $n\pi$

$\tan 3x+\tan x=2\tan 2x\Rightarrow \frac{\sin 3x}{\cos 3x}+\frac{\sin x}{\cos x}=2\frac{\sin 2x}{\cos 2x}\Rightarrow \frac{\sin 4x}{\cos 3x\cos x}=2\frac{\sin 2x}{\cos 2x}\Rightarrow \sin 4x\cos 2x=\sin 2x\left[2\cos 3x\cos x\right]\Rightarrow \sin 4x\cos 2x=\sin 2x[\cos 4x+\cos 2x]\Rightarrow \sin 4x\cos 2x-\sin 2x\cos 4x=\sin 2x\cos 2x\Rightarrow \sin 2x=\sin 2x\cos 2x\Rightarrow \sin 2x=0,\cos 2x=1\Rightarrow x=\frac{n\pi }{2},x=n\pi$
But $x=n\pi /2$ is rejected as when $n$ is odd, $\tan x$ is not defined and when $n$ is even, i.e. $2k$ , then $x=k\pi$
So, $x=n\pi ,n\in Z$ is the solution.

Alternate Solution:
$\tan 3x+\tan x=2\tan 2x\Rightarrow \tan 3x-\tan 2x=\tan 2x-\tan x\Rightarrow \tan x[1+\tan 3x\tan 2x]=\tan x[1+\tan 2x\tan x](\because \tan x=\tan (3x-2x\left)\right)\Rightarrow \tan x\tan 2x[\tan 3x-\tan x]=0\Rightarrow x=n\pi ,x=\frac{n\pi }{2},3x=n\pi +x\Rightarrow x=n\pi ,x=\frac{n\pi }{2}$
But $x=n\pi /2$ is rejected as when $n$ is odd, $\tan x$ is not defined and when $n$ is even, i.e. $2k$ , then $x=k\pi$
So, $x=n\pi ,n\in Z$ is the solution.