Question

How do you solve `2cos^3x + cos^2x = 0` in the interval [0, 2pi]?

Answer 1

`x in {pi/2,(2pi)/3,(4pi)/3,(3pi)/2}`

Explanation:

`2cos^3x+cos^2x=0`
`2cos^2x(cosx+1/2)=0`
`cos^2x=0 or cosx+1/2=0`
`cosx=0 or cosx=-1/2`
`x=pi/2+kpi or (x=(2pi)/3+2kpi or x=(4pi)/3+2kpi)`, where `k in mathbb(Z)`
Those are all real solutions but since we care about only the interval `[0,2pi]` the final solutions are: `pi/2,(2pi)/3,(4pi)/3,(3pi)/2`.