# Thread: Another Pre-Calc Trig Problem

1. ## Another Pre-Calc Trig Problem

given that:
t=sin(18degrees)
what are the steps in solving:
(8t^4)-(8t^2)-t+1=0 ?

2. ## Re: Another Pre-Calc Trig Problem

Hello, jonathanraxa!

I don't understand . . .

Given that: .$\displaystyle t \:=\:\sin(18^o)$
what are the steps in solving: .$\displaystyle 8t^4 - 8t^2 -t+1\:=0\:\,?$

We don't have to solve the equation!

They gave us the answer! .$\displaystyle t \:=\:\sin18^o$

Can you give us the original wording of the problem?

3. ## Re: Another Pre-Calc Trig Problem

The original equation factors out to:

(t-1)(2t+1)(4t^2+2t-1)=0

Since we know t is sin(18), the first two factors will not equal zero, which leaves the final one.

Put that into the quadratic formula and one of the roots you get is (-1+sqrt(5))/4

(-1+sqrt(5))/4=sin(18)

4. ## Re: Another Pre-Calc Trig Problem

I'm plugging sin(18degrees) in for t in the equation.

5. ## Re: Another Pre-Calc Trig Problem

Guess I misunderstood the question.