How do you solve solve cos2x+cos4x=0 ?

Question

anonymous · Answer

Let cos²x = u.

You get u+u² = 0

Factor u out: u(u+1) = 0
u = 0, -1

Now you have cos²x = 0 and cos²x = -1

Does this help you?

anonymous · Answer

I thought you were supposed to use the double-angle properties?

anonymous · Answer

Oh, I though you mean cos²x, sorry. I know you actually mean cos2x. Yes, go ahead and use double-angle properties.

anonymous · Answer

can you explain that? that's what I'm having trouble with

anonymous · Answer

like the first step to solve

anonymous · Answer

We know that $\cos(\alpha + \beta) = \cos\alpha \cos\beta-\sin\alpha\sin\beta$
So cos2x = cos²x - sin²x
cos2x + cos4x = (cos²x - sin²x) + (cos²2x - sin²2x)
                       = (cos²x - sin²x) + ((cos²x - sin²x)² - (2cosxsinx)²)

I honestly don't know if this is correct path to solution, though. Hope I helped.

anonymous · Answer

Also $\sin(\alpha + \beta) = \cos\alpha \sin\beta + \cos\beta\sin\alpha$
So sin2x = cosxsinx + cosxsinx = 2cosxsinx

anonymous · Answer

thank you for your input