SOLUTION: Hi, I have been stuck on this part of a question for a while. I have got to the stage; 4cos^2(36) - 2cos(36) -1 = 0 of a trig. identity but need to simplify this down into the

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Question 847512: Hi, I have been stuck on this part of a question for a while.
I have got to the stage;
4cos^2(36) - 2cos(36) -1 = 0 of a trig. identity but need to simplify this down into the form a+b√5 where a and b are found.
The steps to this confuse me and I don't understand how to do this.
Any help is greatly appreciated.
Thank you!!

Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website!
Hi, I have been stuck on this part of a question for a while.
I have got to the stage;
4cos^2(36) - 2cos(36) -1 = 0 of a trig. identity but need to simplify this down into the form a+b√5 where a and b are found.
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It's not clear what you're trying to do.
That's not an identity. It has no variable.
If it is true, you can solve for cos(36).
Sub x for cos(36)
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Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)

Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=20 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 0.809016994374947, -0.309016994374947. Here's your graph:

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