Question

How do I solve this? cos2X=-cosX on [0, 2pi)

Answer 1

use this :
\[\cos2x=\cos^2x-\sin^2x\]

Answer 2

and this:
\[\sin^2x + \cos^2x=1\]

Answer 3

so you have then:\[\cos2x=\cos^2x-(1-\cos^x)=2\cos^2x-1\]

Answer 4

substitute that in initial equation, sypstitute t= cosx and you will have quadratic equation. Solve for t and then for x

Answer 5

choose only the x's from interval [0, 2pi)

Answer 6

Do you know how to do that?

Answer 7

Yes. I understand how you came to 2cosx^2 -1 but I do not understand after that.

Answer 8

If you substitute that in your initial equation you have:
\[2\cos^2x+cosx-1=0\]

Answer 9

t=cosx and \[t \in<-1;1>\]
so \[2t^2 + t -1=0\]

Answer 10

Do you understand why t must be between -1 and 1?