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
Thanks! I was beginning 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:
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.
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...
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.
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