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cos3q=sinq
By Arun Iyer on Friday, November 23, 2001 - 06:30 pm:

In a particular mechanics problem,
i ran into a equation as such...
cos3q=sinq

is there any easy method of solving this equation....??

i actually solved it by taking cosq=x
which gives...
x3=sqrt(1-x2)
i.e x6+x2-1=0

then i took f(x)=x6+x2-1 and solved for x by Newton Raphson method....

this was quite tedious and time consuming especially during exam....

hence i need any easier method if possible...
love arun

By Yatir Halevi on Friday, November 23, 2001 - 07:16 pm:

actually there is a way,
take x2=m
then we have
m3+m-1=0
And this is the depressed for of the cubic expression (without the x2), and there is a simple way of solving this, described in:
http://www.sosmath.com/algebra/factor/fac11/fac11.html

Hope it helps,
Yatir

By Andrew Hodges on Friday, November 23, 2001 - 07:18 pm:

f(x) is a depressed cubic in x2 if that helps you Arun...

By Kerwin Hui on Saturday, November 24, 2001 - 11:42 am:

Arun,

Here is a quicker way to get to the cubic. First, notice that we cannot have cos q=0, so divide both sides by cos3 q and we obtain

1 = tan q + tan3 q

Now we just plug this into del Ferro's, or use some other methods.

Kerwin

By Arun Iyer on Saturday, November 24, 2001 - 06:04 pm:

Well, thanks for all the methods suggested...

though, I would like to have some quicker method to solve the real equation...i.e
cos3q=sinq

love arun