How do you solve #cos 2theta + 5 cos theta + 3 = 0#?

3 Answers
Abhishek K.
Mar 9, 2018

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

Explanation:

#rarrcos2x+5cosx+3=0#

#rarr2cos^2x-1+5cosx+3=0#

#rarr2cos^2x+5cosx+2=0#

#rarr2cos^2x+4cosx+cosx+2=0#

#rarr2cosx(cosx+2)+1(cosx+2)=0#

#rarr(2cosx+1)(cosx+2)=0#

Either, #2cosx+1=0#

#rarrcosx=-1/2=cos((2pi)/3)#

#rarrx=2npi+-(2pi)/3# where #nrarrZ#

Or, #cosx+2=0#

#rarrcosx=-2# which is unacceptable.

So, the general solution is #x=2npi+-(2pi)/3#.

maganbhai P.
Mar 9, 2018

#theta=2kpi+-(2pi)/3,kinZ#

Explanation:

#cos2theta+5costheta+3=0#
#:.2cos^2theta-1+5costheta+3=0#
#:.2cos^2theta+5costheta+2=0#
#:.2cos^2theta+4costheta+costheta+2=0#
#:.2costheta(costheta+2)+1(costheta+2)=0#
#:.(costheta+2)(2costheta+1)=0#
#=>costheta=-2!in[-1,1],or costheta=-1/2#
#=>costheta=cos(pi-pi/3)=cos((2pi)/3)#
#theta=2kpi+-(2pi)/3,kinZ#

Kalit Gautam
Mar 9, 2018

Use #cos2theta = 2(costheta)^2-1# and the general solution of #costheta =cosalpha# is #theta=2npi+-alpha# ; #n∈Z#

Explanation:

#cos2theta+5costheta+3#

#= 2(costheta)^2-1+5costheta+3#

#= 2(costheta)^2+5costheta+2#

#rArr(costheta+1/2)(costheta+2)=0#

Here #costheta =-2# is not possible

So, we only find the general solutions of #costheta=-1/2#

#rArrcostheta=(2pi)/3#

#:.theta = 2npi+-(2pi)/3 ; n∈Z#

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