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

1. Aug 25, 2006

### sarcaine

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)

2. Aug 25, 2006

### matt grime

tan is sin/cos and cos^2+sin^2=1, that is sufficient to let you solve it.

3. Aug 25, 2006

### sarcaine

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. Aug 25, 2006

### sarcaine

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. Aug 25, 2006

### sarcaine

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. Aug 25, 2006

### matt grime

Why not try to get all sines onto one side of the equation and cosines onto the other?

7. Aug 27, 2006

### sarcaine

How? I still have the sin(x)cos(x) factor.

8. Aug 27, 2006

### Gokul43201

Staff Emeritus
What's wrong with the quartic?

9. Aug 27, 2006

### sarcaine

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. Aug 28, 2006

### sarcaine

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: Aug 28, 2006
11. Aug 28, 2006

### power freak

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!

Your method may lead to some results too, go down whichever route feels comfortable.

12. Aug 28, 2006

### sarcaine

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...

13. Aug 28, 2006

### Staff: Mentor

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. Aug 29, 2006

### sarcaine

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. Sep 9, 2006