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here is the problem cos6x+cos2x=0

sin2xcos2x-cos2x=0

I know you plug in for this I am just not how?

so we use this form in cosucosv=1/2{cosu-v)+cos(u+v0} so now we plug in?

Cos u+Cos v =2Cos((u+v)/2)Cos((u-v)/2)

So now just make the appropriate substitutions for u and v.

when plugging in for U and v does the 1/2 mean a power?

No, it's multiplying the sum and difference of the two angles by 1/2.

cosucosv=1/2 cos 6x+cos2x=0

is this right so far?

Actually, it should be 2cos4xcos2x=0

okay that is what I need help with?

1/2 of 6 is only 3??

Sum of 6x and 2x is 8x divided by 2 is 4x. Difference of 6x and 2x is 4x divided by 2 is 2x.

You do not take halfe of each angle you take half of their sum and then half of their difference.

so you add 6+2= 8 and than divide the sums by 2?

Yes, and then you do it again only use their difference.