Question

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

Answer 1

`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`.

Answer 2

`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`

Answer 3

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`