Question

How would I solve cos x * cos 2x=0 ?

Answer 1

`x=2npi+-pi/2` or `x=npi+-pi/4`

Explanation:

`cosx(cos2x)=0`
`=>" either " cosx=0=>x=2npi+-pi/2 " or "`
`cos2x=0=>2x=2npi+-pi/2" or "x=npi+-pi/4`

Answer 2

`pi/2 + 2kpi; (3pi)/2 + 2kpi`
`pi/4 + kpi; (3pi)/4 + kpi`

Explanation:

cos x.cos 2x = 0
Either factor should be zero.
a. cos x = 0
Unit circle gives -->
`x = pi/2 + 2kpi`, and
`x = (3pi)/2 + 2kpi`
b. cos 2x = 0
Unit circle gives -->
1. `2x = pi/2 + 2kpi`
`x = pi/4 + kpi`
2. `2x = (3pi)/2 + 2kpi`
`x = (3pi)/4 + kpi`