How to solve Asin(x) + Bcos(x) + Ctan(x) = 0

  • Thread starter sarcaine
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  • #1
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I need help in solving this equation

asenx+bcosx+ctanx=0

I solved it a few years ago but now I don't remember the right process.. The solution has to be in this way

Atan^2(x)+Btan(x)+C=0

where A = f(a,b,c), B = g(a,b,c) and C = h(a,b,c)

Thanks in advance
 

Answers and Replies

  • #2
matt grime
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tan is sin/cos and cos^2+sin^2=1, that is sufficient to let you solve it.
 
  • #3
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Yes, these are the formulas I'm using. But I need some more help.. please show me some of the steps I can follow to solve it
 
  • #4
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I'm trying:

asin(x)+bcos(x)=-ctan(x)
a^2sin^2(x)+2absin(x)cos(x)+b^2cos^2(x)=c^2sin^2(x)/cos^2(x)

I know I have to quit the sin(x)cos(x) term, how to do it? Please, give me some ideas, I'll thank you a lot.
 
  • #5
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Other way I've tried is this
we call
x = tan^(-1)(y)
then
ay/sqrt(1+y^2)+b/sqrt(1+y^2)+cy=0
ay+b+cy*sqrt(1+y^2)=0
ay+b=-cy*sqrt(1+y^2)
a^2y^2+2aby+b^2=c^2y^2(1+y^2)
a^2y^2+2aby+b^2=c^2y^2+c^2y^4
c^2y^4+(c^2-a^2)y^2-2aby-b^2=0

The result is a quartic equation where y=tan(x), but this is not the solution I want to get.
 
  • #6
matt grime
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Why not try to get all sines onto one side of the equation and cosines onto the other?
 
  • #7
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How? I still have the sin(x)cos(x) factor.
 
  • #8
Gokul43201
Staff Emeritus
Science Advisor
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sarcaine said:
The result is a quartic equation where y=tan(x), but this is not the solution I want to get.
What's wrong with the quartic?
 
  • #9
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It's just that this equation can be expressed in this way

Atan^2(x)+Btan(x)+C=0

This is much more pretty than the quartic.
Other reason is that t's not easy to solve the quartic because of the 2aby factor.
 
  • #10
11
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I've reached this equation

(a^2+b^2)-2cb/sin(x)-b^2/sin^2(x)=c^2/cos^2(x)

Don't know next step
 
Last edited:
  • #11
sarcaine said:
a^2sin^2(x)+2absin(x)cos(x)+b^2cos^2(x)=c^2sin^2(x)/cos^2(x)
I've solved the problem the way you asked me too, I'm not going to step you through it as that ruins the fun...

I basically started from this point though, can you do anything to the right hand side that would cancel a few things on the left hand side?

To use my method you'll have to adapt the sin^2 + cos^2 = 1 equation ever so slightly.

:)

From the point you got to in the quote above there's only 2 lines of working left!


:cool:

Your method may lead to some results too, go down whichever route feels comfortable.
 
  • #12
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Thanks! I was begining to believe that it was a wrong method. Now I know it can be solved that way.

And I remember I did something like what you're saying , if you could lend me some more information...

:smile:
 
  • #13
berkeman
Mentor
60,860
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For me, it helps to look at the graphs of sin, cos and tan at the beginning of a problem like this, in order to understand what the solution(s) will be, and to understand how to limit the domain of the solution. Maybe just try it briefly to see if it lends some intuition to what you are doing.

Just plot sin, cos and tan over a couple of cycles left and right of the origin, including the asymptotes of the tan function. Do you see some angles where you can get the three to add up to zero, given some A B and C values?
 
  • #14
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power freak said:
To use my method you'll have to adapt the sin^2 + cos^2 = 1 equation ever so slightly.

:)

From the point you got to in the quote above there's only 2 lines of working left!

I'm trying to get it this way but I can't reach it. Could you be more specific? What does it mean slightly? Do I have to add 1 to both sides of the equation? Thanks
 
  • #15
11
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please help!
 
  • #16
11
0
I need help!
 

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