Trigonometric Roots and Substitutions: Solving Equations in Terms of Pi

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The discussion revolves around solving the cubic equation 2x^3-5x^2-4x+3=0, with initial roots incorrectly identified as x=-1, 3, and 1. The correct roots are identified as x=-1, 3, and 1/2. The second part of the problem involves substituting x=cos(t) into the equation, leading to the need to solve for cos(t)=-1 and cos(t)=1/2. This results in multiple solutions for t, with the possibility of complex solutions if cos(t)=3 is considered. The conversation highlights the importance of accurately finding roots and understanding trigonometric identities in solving equations.
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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.

PLEASE HELP! :confused: :confused: :confused:
 
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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:
Gughanath said:
giving your answer in radians in terms of pie.

PLEASE HELP!

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 :biggrin:
 

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