Find all solutions to equation in interval [0,2pi) ? cos^2x+cosx = cos 2x

interval-02pi

#1

Find all solutions to equation in interval [0,2pi) ? cos^2x+cosx = cos 2x

Answer:

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cos 2x=2*cos^2 x-1
so:
cos^2x+cosx = cos 2x
<=>cos^2x-cosx-1=0
take u=cos x
then we have to solve:
u^2-u-1=0
This gives: u=(1±sqrt(5))/2
but (1+sqrt(5))/2>1 and therefor isn’t possible.
So we find:
cos x= (1-sqrt(5))/2
so x=2,237 or4,046