How do you solve #cos 2θ + cosθ = 0# from #[0,2pi]#?

1 Answer
Sasha P.
Sep 15, 2015

#theta=pi/3 vv theta=(5pi)/3 vv theta=pi#

Explanation:

#cos2theta=cos^2theta-sin^2theta#
#cos^2theta-sin^2theta+costheta=0#
#sin^2theta+cos^2theta=1 => sin^2theta=1-cos^2theta#
#cos^2theta-(1-cos^2theta)+costheta=0#
#cos^2theta-1+cos^2theta+costheta=0#
#2cos^2theta+costheta-1=0#
#2cos^2theta-costheta+2costheta-1=0#
#costheta(2costheta-1)+2costheta-1=0#
#(2costheta-1)(costheta+1)=0#
#2costheta-1=0 vv costheta+1=0#
#costheta=1/2 vv costheta=-1#
#theta=pi/3 vv theta=(5pi)/3 vv theta=pi#

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