I got the answer for the process cosx+cos2x=0, but i got lost in a step.?

Question

this transforms the above to equation to p+2p^2-1=0
This factors as (2p-1)(p+1)=0

¿Can I replace in all the equations with p?

Answer 1

There is a trig identity you need: cos(2x) = 2cos^2(x)-1.
If you want to derive it, you can use the cosine of sum formula:
cos(x+x) = cos(x)cos(x)-sin(x)sin(x) and go from there...

You can apply this to your equation cosx+cos(2x)=0

This gives cos(x)+2cos^2(x)-1 = 0

Then yes, you can substitute p for cos(x) giving

2p^2 + p - 1 = 0

(2p-1)(p+1) = 0

p = 1/2, p = -1

cos(x) = 1/2, -1

x = Arccos(1/2) or Arccos(-1)

x = 60 degrees or 180 degrees

Answer 2

Yes. p is identical to cosx, and your eq in p is correctly derived, so by inspection, p = 0.5 or -1----> cosx = 0.5 or -1----> x = +-60 deg or +- 180 deg

Answer 3

I suspect that, like myself, most of the readers don't understand what you want.

Try to explain....