Question

General solution of $\tan x+\tan 2x=0$ is

• A. $n\pi ,n\pi \pm \frac{\pi }{3},n\in Z$
• B. $\frac{n\pi }{2},n\pi \pm {\tan }^{-1}\left(\frac{1}{\sqrt{2}}\right),n\in Z$
• C. $\frac{n\pi }{3},n\pi \pm {\tan }^{-1}\left(\frac{1}{2}\right),n\in Z$
• D. $\frac{n\pi }{3},n\pi \pm {\tan }^{-1}\left(\frac{1}{\sqrt{2}}\right),n\in Z$

Answer 1

The correct option is A $n\pi ,n\pi \pm \frac{\pi }{3},n\in Z$

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

$⟹\tan x+\tan 2x=0$

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

$⟹\tan x(1+\frac{2}{1-{\tan }^{2}x})=0$

$⟹\tan x\times \frac{1-{\tan }^{2}x+2}{1-{\tan }^{2}x}=0$

$⟹\tan x(3-{\tan }^{2}x)=0$

$⟹\tan x=0$ or ${\tan }^{2}x=3$

$⟹x=n\pi$ or $x=n\pi \pm \frac{\pi }{3}$