Hi there

 

If i had a function in the format

cos(ln(x+1)) =0

how would i go about trying to solve for the solutions (roots)

I can see there are three roots but i'm not sure how to solve it.

 

ps. how do you solve secx(2sec^2x-1)=0

as ih happens i know it has no solutions, but how would i go about showing this. where can i look this stuff up

 

thanks guys!

Your first one looks like it would boil down to x=e^arccos(0)-1. There should be an infinite number of solutions since the cos(x)=0 at every point pi/2+npi where n is an integer.

Hint for the second one: If a*b=0 then one of a or be is zero.

if: cos(ln(x+1)) = 0

 

use: 0 = cos(pi/2 + n*pi)

 

cos(ln(x+1)) = cos(pi/2 + n*pi)

ln(x+1) = pi/2 + n*pi

x+1 = e^(pi/2 + n*pi)

 

therefor: x = e^(pi/2 + n*pi) - 1

 

I believe that's it. But correct me if I'm wrong. I'm just a poor biochemist who tried to escape math a long time ago

Seems to be okay to me. You don't necessarily need the cos(stuff) = cos(bleh) line.

thanks guys!!!

I only just realised that i had to treat the two parts cosx=o and ln(x+1) as separate functions. Sorry i'm new to this type of stuff.

I find problems with solving a function that is not equal to 0

ie. secx(2secx^2-1)=-1

 

what do i do with this beauty to find its solutions? is it secx=-1 and secx=sqrt-2 ,which is impossible?

Yeah, you can only set the factors equal to the other side when the other side is zero. That's because only zero has the identity 0X=0 for all X's. So I would start by moving the secx to the other side which gives -1/secx, or -cosx. So now the equation is 2sec²x-1=-cosx or 2sec²x=1-cosx if you prefer. From there, I would look through yourdad's thread on trig identities and see what you can come up through manipulation:

http://www.scienceforums.net/forums/showthread.php?t=9608

cheers for the help!

 

so

secx(2sec^2-1)=-1

=> 2sec^2-1 =-cosx 'using tan^2x+1=sec^2

2(1+tan^2x)-1=-cosx

2+2sin^2x/cos^2x=-cosx 'using tan^2x=sin^2x/cos^2x

2(cos^2x+sin^2x)=-cos^2x ' using c^2+s^2=1

-2=cos^3x

 

that seems to be it. but i don't see how it could be that. if between -pi and pi, then there would be toooo many solutions! hundreds even! no?

2(1+tan^2x)-1=-cosx

2+2sin^2x/cos^2x=-cosx

 

There seems to be a mistake in this step. It should be 2+2tan²x-1 which equals 1+2tan²x.

sorry i don't follow

if i'm using tan^2x+1=sec^2x

why wouldn't it be 2(sec^2x)=2(tan^2x+1)?

Well, let's just rearange this line for now:

 

2(1+tan^2x)-1=-cosx

 

First thing is to distribute the 2:

 

(2+2tan²x)-1=-cosx

 

Then get rid of the parenthesis and subtract the 1 from the 2. That gives:

 

1+2tan²x=-cosx

 

Use the identity tan²x=sin²x/cos²x to get:

 

1+2sin²x/cos²x=-cosx

 

That's just a little different than your line:

 

2+2sin²x/cos²x=-cosx

Ok i understand now.

so if we were to work it through in the same manner:

1+2sin^2x/cos^2x=-cosx

 

2sin^2x+cos^2x=-cos^3x

2(-cos^2x+1)=+cos^2x=-cos^3x

 

2-2cos^2x=-cos^3x

2-cos^2x=-cos^3x

 

and where do i go from here?

2(-cos^2x+1)=+cos^2x=-cos^3x

2-2cos^2x=-cos^3x

I think there's another algebra mistake here.

Let's distribute the 2 again:

 

(2-2cos²x)+cos²x=cos³x

 

It looks like you forgot the lone cos² on the left side (maybe because of the errant equals sign?)

 

Either way' date=' that boilds down to:

 

2-cos[sup']2[/sup]x=cos³x

 

2=cos³x+cos²x

Really, I wish I could be more help, but I don't this is the point where I origionaly had it and couldn't go much further for some reason. You seem to find some pretty clever ways to manipulate it though, so keep posting what you try.

I find problems with solving a function that is not equal to 0

ie. secx(2secx^2-1)=-1

 

what do i do with this beauty to find its solutions? is it secx=-1 and secx=sqrt-2 ' date='which is impossible?[/quote']

 

We can treat this as a polynomial in secx, as I think you were hinting at above.

 

Letting u = secx, consider

f(u) = u(2u² - 1) + 1

= 2u³ - u + 1

= (u + 1)(2u² - 2u + 1)

 

f(u) = 0 => u + 1 = 0 or 2u² - 2u + 1 = 0.

From the linear factor we have u = -1.

The quadratic factor has no real roots, as its discriminant is negative.

 

Hence the only solutions are where u = secx = -1.

That is, cosx = -1, so that x = (2n + 1)*pi, where n is an integer.

Ah...yep. That should work too. Nice work there.

cheers! thanks a lot!

 

my problem was that i didn't know if the way i was trying to solve the function was the correct one.

 

thanks again