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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