Question

TUTORIAL :
on how to solve a 3 degree eqn in x using trigonometry .. i learnt this method quite some time ago and maybe everyone should give this a reading..

Answer 1

we have :

ax^3 + bx^2 + cx+ d =0

for simplification, i like to keep coefficient of x^3 = 1
for that,,lets divide both sides by a (given a not equal to 0)

lets say b/a =p , c/a = q , d/a =r

=> x^3 + px^2 + qx +r =0

what we aim here firstly is to convert this into depressed form ( make co-efficient of 2nd highest degree =0)

lets substitute x = (y + m) , for some new variable y and a constant m , we will understand the purpose of doing this soon..

=> (y+m)^3 + p(y+m)^2 + q(y+m) + r =0
=>y^3 + y^2(p + 3m) + y(2m + 3m^2 +q) + (m^3 + pm^2 + qm +r) =0

as i said earlier,we want
p+ 3m =0
=> m= -p/3

thus on making that substitution, we'll get

y^3 + Ay + B = 0

where A and B are some known constants..
now if we calculate y , we can easily calculate x as x= y+m or x= y -p/3

here comes the trignometric part:

this glorious formulla will do wonders for us as you'll find below :
sin 3@ = 3sin @ - 4sin^3 @

(@ denotes theta)

so we have
y^3 + Ay + B=0

let y = t sin@ for some constant t ,

=> t^3 sin^3 @ + At sin@ + B =0
lets multiply both sides by -4/t^3

=> -4 sin^3 @ + -4A sin@ /t^2 + -4B/t^3 = 0

we have to chose t such that coefficient of sin@ = 3
=> -4A/t^2 = 3
=>t = sqrt( -4A/3) and -sqrt( -4A/3)

on making that substitution, we see we have

-4sin^3 @ + 3sin@ =C
for some new constant C.
=> sin3@ = C
=>3@ = sin^-1 (C)
@ = 1/3 * sin^-1 (C)

since you know the value of @, you then know value of y = t sin@
and then x = y+m

Answer 2

all this might seem tedious and long at first,,but this is very advantageous to know..
this one yields both complex roots as well as real roots..2 of them .. (since you get 2 values of t)

hope that helps..

Answer 3

another thing which we can make out from here that, in any eqn like

x^n + px^(n-1) + qx^(n-2) ......... + constant term =0

to make it depressed eqn , we can always substitute x = y - (p/n)

Answer 4

What is this method called ?

Answer 5

i dont know/remember.. :|

Answer 6

k

Answer 7

3x^3+2x^2+x+1=0

Show me how to solve this step by step please.

Answer 8

you do it and i shall add my words in between when required..

Answer 9

let me start then :)

Answer 10

Do i have to suppose x^3=1 ?

Answer 11

nops..i meant coefficient of x^3 should be made 1,,just for simplification..
so you may divide both sides by 3

Answer 12

Dividing the whole equation by 3

x^3+2/3 x^2+1/3x+1/3=0

Answer 13

What next ?

Answer 14

x = y - p/3 .. make that substitution,,
p = 2/3 here..

Answer 15

p=2/3,q=1/3,r=1/3

I did not get the substitution part lol ?

Do you think this method is more effective then the others ?

Answer 16

i cant say that,,since i dont know other methods! :P
nevermind,,whats the problem is substituting ?

Answer 17

like in x+2 = y, you can substitute directly x=2 ,, similarly,,substitute x= y - p/3

Answer 18

i'll give a kick..

x^3+2/3 x^2+1/3x+1/3=0

(y - 2/9)^3+ 2/3 (y- 2/9)^3 + 1/3 ( y- 2/9) + 1/3 = 0

did you follow?

Answer 19

Ok so we have just done a simple substitution here x=y-2/9.

Answer 20

Do i expand them using formula ?

Answer 21

yep..expand and simplify ..
write what you get .. sorry am too lazy to check your work or do it myself..so be careful..please dont mind. :P

Answer 22

Anyways i understood the part where i was confused,Thanks a lot.

Answer 23

so,,you're comfortable with the rest ?

Answer 24

Yeah i was confused in this part lol :D

Answer 25

kk.glad i could help..

Answer 26

Is there a textbook or video tutorial I can find this in?

Answer 27

If i remember the name of the method i could have told you :(

Answer 28

well one of my seniors had told me this..i really cant tell you any name or any vids or theories about it..sorry..

anyways,,are you stuck somewhere ?

Answer 29

@Hero This method does not intrest me :) Its very long.

Answer 30

sorry buddy,,too lazy doing all that..though i could help you ..but giving examples..nah,,not gonna do that..

Answer 31

ZZZZZzzz...

Answer 32

@shubhamsrg Please.

Answer 33

i dont regret that..i made an effort..and it was satisfactory for me..i dont mind your foolish reviews since you ofcorse,dont get it..

Answer 34

@shubhamsrg If you give an example maybe you would be getting more medals think about that :D

Answer 35

ohh yes,,the medals..it was always my dream to earn more and more medals on open study.. !! o.O

Answer 36

Oh come on just provide us with an example please ?

Answer 37

If you feel lazy do it after sometime :)

Answer 38

nevermind,,not gonna do that..if you dont get it..put up your query here..i may help/improvise// but i hate doing hard work more often than not..

Answer 39

it is tooo lengthy
but indeed beautiful
we use to call it 'MAJDOORI'

Answer 40

we?? o.O

who all do you represent? :P

Answer 41

alright,,seems its lengthy..

in that case,,gimme some time,,i'll post a general formula to the value of x directly..

Answer 42

except for the trigonometric substitution, solving a cubic equation appeared in Ars Magna of Girolamo Cardamo, or Cardan.

Answer 43

here's the general formula for solution of x in x^3 + px^2 + qx + r =0 using trig. substitution :