Guest

Q.1 tanx + tan 2x + tan 3x = 0 Q.2 sec x - 1 = ( 1.414 - 1) tanx Ans. 1. n3.14/3, [ n3.14 (+-) tan inverse 1/1.414] ans 2. 2n3.14 + 3.14/4

Q.1 tanx + tan 2x + tan 3x = 0


Q.2 sec x - 1 = ( 1.414 - 1) tanx


Ans. 1.   n3.14/3, [ n3.14  (+-)  tan inverse 1/1.414]


ans 2.   2n3.14 + 3.14/4

Grade:

2 Answers

SAGAR SINGH - IIT DELHI
879 Points
12 years ago

Dear student

tanx+tan2x+tan3x=0

(tanx+tan2x)(1-tanxtan2x)/((1-tanxtan2x)+tan3x=0

tan3x(1-tanxtan2x)+tan3x=0

so we get

either tan3x=0 ortanxtan2x=2

Solve for x....

Please feel free to ask your queries here. We are all IITians and here to help you in your IIT JEE preparation.

All the best.

Win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian.

Now you score 5+15 POINTS by uploading your Pic and Downloading the Askiitians Toolbar  respectively : Click here to download the toolbar..

 

Askiitians Expert

Sagar Singh

B.Tech, IIT Delhi

sagarsingh24.iitd@gmail.com

Harsha Kumari
14 Points
4 years ago
 
tanx + tan2x+ tan3x=0
or, tanx+ tan2x= - tan3x
or, (sinx/cosx)+(sin 2x /Cos 2x)=-(sin 3x /cos 3x)
or, (sin x *cos 2 X + cos x *sin 2x)/cos x * cos 2x = -sin 3x / cos 3x
or, sin( 2x+x)* cos 3x= - cos x *cos 2x *sin 3x
or, sin 3x*cos 3x+ cos x * cos 2x *sin 3x =0
or, sin 3x ( cos 3x + cos x *cos 2x)=0
or, sin 3x (cos ( 2x + x)+ cos x * cos 2x)=0
or, sin 3x (cos x *cos 2x - sin x *sin 2x+ cos x *cos 2x) =0
or, -sin 3x *sin x * sin 2x=0
 
Either sin3x=0
i.e, 3x=nπ
i.e, X=nπ/3
 
or, sin 2x=0
i.e, 2x=nπ
i.e, X= nπ/2
 
or, sin x= 0
i.e, X=nπ
Required solution is X= nπ/3

Think You Can Provide A Better Answer ?