Question

Find the no. of solutions satisfying the equation-

Answer 1

\[\tan^{2}2x=2\tan2x.\tan3x+1\] where x lies in [0,2pi]

Answer 2

converting everything to sines and cosines might help

Answer 3

means x is in 3rd quadrant?

Answer 4

i want sol.

Answer 5

i love trig lets try.

Answer 6

tan^22x? please check

Answer 7

its correct

Answer 8

sorry - ill like to have a go at this one but i gotta go

Answer 9

confusing me.

Answer 10

can we try opening to sin and cos

Answer 11

you can do anything

Answer 12

\[-1=2(\frac{ \sin2x }{ \cos2x })(\frac{ \sin3x }{ \cos3x })-\frac{ \sin^2 2x}{ \cos^2 2x }\]
\[-\frac{ 1 }{ 2 }=(\frac{ \sin6x }{ \cos6x })-\frac{ \sin^2 2x}{ \cos^2 2x}\]
\[-\frac{ 1 }{ 2 }=\tan6x-\tan^22x\]
and im lost

Answer 13

@hartnn

Answer 14

i can think of this:
tan x tan 2x tan 3x = tan x+ tan 2x + tan 3x
but i don't know how to use it....just a thought.

Answer 15

x+2x=3x
tan(x+2x)=tan3x
(tanx+tan2x)/(1-tanx * tan2x)=tan3x
tanx+tan 2x =tan3x -tanx * tan2x * tan3x
tanx* tan2x * tan3x = tan3x - tan2x - tanx
@hartnn how did u get that?