Question

How do you solve `2cos^2x+3cosx-2=0`?

Answer 1

Let `n=cosx`

`2n^2+3n-2=0`

Solve for n by factoring:
`2n^2-n+4n-2=0`
`n(2n-1)+2(2n-1)=0`
`(n+2)(2n-1)=0`

`n+2=0`
`n=-2`

`2n-1=0`
`n=1/2`

Substitute `n=cosx` in for those two answers:
`cosxcancel(=)-2`
Cosine can never equal -2.

`cosx=1/2`

`x=pi/3, (5pi)/3`