Question

How do I solve for 0º ≤ x < 360º using this equation 2 cos² x + 3 cos x -2 = 0 ?

Answer 1

`x=pi/3+2kpi`
`x=-pi/3+2kpi`

Explanation:

`2cos^2x+3cos-2=0`

`sqrt(∆)=sqrt(25)=5`
`t_1=(-3-5)/4=-2`
`t_2=(-3+5)/4=1/2`

`cosx=1/2`

`x=pi/3+2kpi`
`x=-pi/3+2kpi`

k is real