Solutions to equation in interval [0, 2pi)

[Chaim]

cos 2x = cos x

Well first thing I would have to do is convert this into a quadratic formula (meaning ax²+bx+c)
Though, my teacher helped me get started, and told me to do this:

2cos²(x)-cos(x)-1
I was lost at how the cos 2(x) became 2cos²x

Can anyone explain to me this?

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Though, after that I would factor it then find the x's from there.

 

[Deleted member 4993]

cos 2x = cos x

Well first thing I would have to do is convert this into a quadratic formula (meaning ax²+bx+c)
Though, my teacher helped me get started, and told me to do this:

2cos²(x)-cos(x)-1
I was lost at how the cos 2(x) became 2cos²x

Can anyone explain to me this?

----
Though, after that I would factor it then find the x's from there.

cos(2x) = cos²x - sin²x = 2*cos²x - 1

 

[Chaim]

cos(2x) = cos²x - sin²x = 2*cos²x - 1

Ah!
This is the Pythagorean Identities right?
sin²θ+cos²θ=1

Ok thanks!

 

[Deleted member 4993]

Ah!
This is the Pythagorean Identities right?
sin²θ+cos²θ=1

Ok thanks!

No not really - however we do use the identity on the way.

First we use the angle-sum identity:

cos(A+B) = cos(A)*cos(B) - sin(A)*sin(B) → cos(X+X) = cos²(X) - sin²(X)

Now we use Pythagorean identity to get to last part....

 

[Chaim]

No not really - however we do use the identity on the way.

First we use the angle-sum identity:

cos(A+B) = cos(A)*cos(B) - sin(A)*sin(B) → cos(X+X) = cos²(X) - sin²(X)

Now we use Pythagorean identity to get to last part....

Ah I see!
Thank you very much!