How would I solve #cos x + cos 2x = 0#? Please show steps.

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Answer 1 · 2015-11-22T21:31:37

We know that

#cos2x=cos^2x-sin^2x=cos^2x-(1-cos^2x)=2cos^2x-1#

hence the equation is

#cosx+cos2x=cosx+2cos^2x-1#

Hence we have to solve the

#2cos^2x+cosx-1=0=>(cosx+1)*(2cosx-1)=0#

or

#cosx=-1=>cosx=cospi=>x=2*k*pi+-pi#

and

#2cosx-1=0=>cosx=1/2=>cosx=cos(pi/3)=>x=2*k*pi+-pi/3#