# Triginometry - very annoying

1. May 1, 2005

### Gughanath

The first part of this question was to find the roots of the equation 2x^3-5x^2-4x+3=0 i got x=-1, 3, and 1.

but then the second part completely confused me
b) Hence, by substituting x=cost solve the equation 2cos^3t-5cos^2t -4cost+3=0 for 0<t<2"pie", giving your answer in radians in terms of pie.

2. May 1, 2005

### Zurtex

Can you solve cos(t)=-1?

Oh, you've solve the cubic equation wrong, 1 of the solutions for x does not work.

Last edited: May 1, 2005
3. May 1, 2005

### kleinwolf

The roots are : x=-1,3,1/2.

Just solve : cos(t)=-1 and cos(t)=1/2...this gives of course an infinity of real answers...

If you allow complex numbers, then you can solve cos(t)=3...

NB : You're great, with your help I know how the english pronounciation of $$\pi$$ double explain it's origin