Question

How do you solve `2cosx+1=0`?

Answer 1

`S={x|x=2/3pi+2kpivvx=4/3pi+2kpi " where " kinZZ}`

Explanation:

`2cosx+1=0`
`cosx=-1/2`
We remember the values for which `cosx=-1/2`.
`x=2/3pi=120° or x=4/3pi=240°.`

But don't forget that cos(x) has a `2pi` periodicity, so:
`S={x|x=2/3pi+2kpivvx=4/3pi+2kpi " where " kinZZ}`