# Solve the following equations:

Question:

Solve the following equations:

(i) $\tan x+\tan 2 x+\tan 3 x=0$

(ii) $\tan x+\tan 2 x=\tan 3 x$

(iii) $\tan 3 x+\tan x=2 \tan 2 x$

Solution:

(i) We have: $\tan x+\tan 2 x+\tan 3 x=0$

Now,

$\tan x+\tan 2 x+\tan (x+2 x)=0$

$\Rightarrow \tan x+\tan 2 x+\left(\frac{\tan x+\tan 2 x}{1-\tan x \tan 2 x}\right)=0$

$\Rightarrow(\tan x+\tan 2 x)(1-\tan x \tan 2 x)+\tan x+\tan 2 x=0$

$\Rightarrow(\tan x+\tan 2 x)(2-\tan x \tan 2 x)=0$

$\Rightarrow \tan x+\tan 2 x=0$ or $2-\tan x \tan 2 x=0$

Now,

$\tan x+\tan 2 \mathrm{x}=0$

$\Rightarrow \tan x=-\tan 2 x$

$\Rightarrow \tan x=\tan -2 x$

$\Rightarrow x=n \pi-2 x$

$\Rightarrow 3 x=n \pi$

$\Rightarrow x=\frac{n \pi}{3}, n \in Z$

And,

$2-\tan x \tan 2 x=0$

$\Rightarrow \tan x \tan 2 x=2$

$\Rightarrow \frac{\sin x}{\cos x} \frac{\sin 2 x}{\cos 2 x}=2$

$\Rightarrow \frac{2 \sin ^{2} x \cos x}{\cos x}=2 \cos ^{2} x-2 \sin ^{2} x$

$\Rightarrow 4 \sin ^{2} x=2 \cos ^{2} x$

$\Rightarrow \tan ^{2} x=\frac{1}{2} \Rightarrow \tan ^{2} x=\tan ^{2} \alpha$

$\Rightarrow x=m \pi+\alpha, m \in Z, \alpha=\tan ^{-1}\left(\frac{1}{2}\right)$

$\therefore x=\frac{n \pi}{3}, n \in Z$ or $x=m \pi+\alpha, m \in Z$

here,

$\alpha=\tan ^{-1}\left(\frac{1}{2}\right)$

(ii) Given:

$\tan x+\tan 2 x=\tan 3 x$

Now,

$\tan x+\tan 2 x=\tan (x+2 x)$

$\Rightarrow \tan x+\tan 2 x=\frac{\tan x+\tan 2 x}{1-\tan x \tan 2 x}$

$\Rightarrow \tan x+\tan 2 x-\frac{\tan x+\tan 2 x}{1-\tan x \tan 2 x}=0$

$\Rightarrow(\tan x+\tan 2 x)(1-\tan x \tan 2 x)-(\tan x+\tan 2 x)=0$

$\Rightarrow(\tan x+\tan 2 x)(1-\tan x \tan 2 x-1)=0$

$\Rightarrow(\tan x+\tan 2 x)(-\tan x \tan 2 x)=0$

$\Rightarrow \tan x+\tan 2 x=0$ or $\tan x \tan 2 x=0$

Now,

$\tan x+\tan 2 x=0$

$\Rightarrow \tan x=-\tan 2 x$

$\Rightarrow \tan x=\tan -2 x$

$\Rightarrow x=n \pi-2 x, \quad n \in Z$

$\Rightarrow 3 x=n \pi$

$\Rightarrow x=\frac{n \pi}{3}, n \in Z$

And,

$\tan x \tan 2 x=0$

$\Rightarrow \frac{\sin x}{\cos x} \frac{\sin 2 x}{\cos 2 x}=0$

$\Rightarrow \frac{2 \sin ^{2} x}{\cos ^{2} x-\sin ^{2} x}=0$

$\Rightarrow \sin ^{2} x=0 \Rightarrow \sin ^{2} x=\sin ^{2} 0$

$\Rightarrow x=m \pi, m \in Z$

$\therefore x=\frac{n \pi}{3}, n \in Z$ or $x=m \pi, m \in Z$

(iii) Given: $\tan 3 x+\tan x=2 \tan 2 x$

Now,

$\tan 3 x-\tan 2 x=\tan 2 x-\tan x$

$\Rightarrow \tan x(1+\tan 3 x \tan 2 x)=\tan x(1+\tan 2 x \tan x)$

$\left[\tan (A-B)=\frac{\tan A-\tan B}{1+\tan A \tan B}\right]$

$\Rightarrow \tan x(1+\tan 3 x \tan 2 x-1-\tan 2 x \tan x)=0$

$\Rightarrow \tan x \tan 2 x(\tan 3 x-\tan x)=0$

$\Rightarrow \tan 2 x=0$ or, $\tan x=0$ or, $\tan 3 x-\tan x=0$

And, $\tan 2 x=0 \Rightarrow 2 x=n \pi \Rightarrow x=\frac{n \pi}{2}, n \in Z$

Or, $\tan 3 x-\tan x=0 \Rightarrow \tan 3 x=\tan x \Rightarrow 3 x=n \pi+x \Rightarrow 2 x=n \pi \Rightarrow \mathrm{x}=\frac{\mathrm{n} \pi}{2}, \mathrm{n} \in \mathrm{Z}$

And, $\tan x=0 \Rightarrow x=m \pi, m \in Z$

$\therefore x=\frac{n \pi}{2}, n \in Z$ or $x=m \pi, \mathrm{m} \in \mathrm{Z}$