Please help solve... (1 Viewer)

tommykins

i am number -e^i*pi
Joined
Feb 18, 2007
Messages
5,730
Gender
Male
HSC
2008
cos^4x = tan^2x
cos^4 x = sec^2x - 1
cos^6 x = 1 - cos^2x

solve that.
 

study-freak

Bored of
Joined
Feb 8, 2008
Messages
1,133
Gender
Male
HSC
2009
If anyone can solve..
cos^2 x=tan x
1=tan x sec^2 x
1=tan x (1+tan^2 x)
tan^3 x + tan x -1=0

I just can't solve after this..
 

study-freak

Bored of
Joined
Feb 8, 2008
Messages
1,133
Gender
Male
HSC
2009
OH I got it... I'll post it soon (I hope it's correct.)
Edit: Whoops. it was wrong... Still can't do it..
 
Last edited:

study-freak

Bored of
Joined
Feb 8, 2008
Messages
1,133
Gender
Male
HSC
2009
lol, approximation of roots
LOL....

Anyway, since I'm getting something imaginary, can someone plz correct my working...

From my previous post in this thread,
tan^3 x+tan x -1=0
Differentiating in terms of x,
3tan^2 x sec^2 x+sec^2 x=0
3tan^2 x=-1
tan^2 x=-1/3
tan x=i/root3
wth... imaginary number...
 

lyounamu

Reborn
Joined
Oct 28, 2007
Messages
9,998
Gender
Male
HSC
N/A
LOL....

Anyway, since I'm getting something imaginary, can someone plz correct my working...

From my previous post in this thread,
tan^3 x+tan x -1=0
Differentiating in terms of x,
3tan^2 x sec^2 x+sec^2 x=0
3tan^2 x=-1
tan^2 x=-1/3
tan x=i/root3
wth... imaginary number...
lol no,,, you cannot get imaginary because i clearly got some angles there.
 

tommykins

i am number -e^i*pi
Joined
Feb 18, 2007
Messages
5,730
Gender
Male
HSC
2008
that just says theres no tp's or inflexion.
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top