Question

How do you solve `1+ cos x + cos 2x = 0`?

Answer 1

`x=pi/2+kpi,(2pi)/3+2kpi,(4pi)/3+2kpi,kinZZ`

Explanation:

Use the identity for the cosine double angle formula:

`cos2x=2cos^2x-1`

The equation then becomes

`1+cosx+2cos^2x-1=0`

`2cos^2x+cosx=0`

Factor out a `cosx` term.

`cosx(2cosx+1)=0`

Set each of these equal to `0`.

`cosx=0" "" ""and"" "" "2cosx+1=0`
`" "" "" "" "" "" "" "" "" "cosx=-1/2`

`cosx=0` occurs at `x=pi/2,(3pi)/2,(5pi)/2`, which can be generalized as `x=pi/2+kpi,kinZZ` (this means where `k` is an integer).

`cosx=-1/2` occurs at `x=(2pi)/3,(4pi)/3` and all other concentric angles (which are `2pi` away), thus we have `x=(2pi)/3+2kpi` and `x=(4pi)/3+2kpi`, both where `kinZZ`.