How do you solve #cos(2(theta)) + 5 cos (theta) + 3#?

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Answer 1 · 2017-02-02T07:25:48

#theta=2npi+-(2pi)/3#

It appears that the questioner meant

#cos2theta+5costheta+3=0#

As this is equiavalent to #2cos^2theta-1+5costheta+3=0#

or #2cos^2theta+5costheta+2=0#

and using quadratic formula

#costheta=(-5+-sqrt(5^2-4xx2xx2))/4=(-5+-3)/4#

i.e #costheta=-2# or #-1/2#

But as range of #costheta# is #[-1,1]#, we cannot have #costheta=-2#

Therefore #costheta=-1/2=cos((2pi)/3)#

Hence #theta=2npi+-(2pi)/3#