Question

Question #4d136

Answer 1

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

Explanation:

`cos^2(x)+cosx=0`

factor out `cos(x)`

`cos(x)[cos(x)+1]=0`

Use the zero property

`cos(x)=0` equation 1
`cos(x)+1=0` equation 2

Equation 1: `cos(x)=0` when x = `pi/2` and `(3pi)/2`

Equation 2: `cos(x)=-1` when x = `pi`

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

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