Question

How do you solve `cos^2x+cos x=0` for `0<=x<=2pi`?

Answer 1

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

Explanation:

Given
`color(white)("XXX")cos^2(x)+cos(x)=0`

Factoring:
`color(white)("XXX")cos(x) * (cos(x)+1)=0`

`rArr` (for `x in [0,2pi]`)
#color(white)("XXXXX"){: (cos(x)=0,color(white)("XXX"),cos(x)+1=0), (x=pi/2 or x=(3pi)/2,,cos(x)=-1), (,,x=pi) :}#