Question

How do you solve `2cos^2x=1`?

Answer 1

`2cos^2x=1`

`=>cosx=+-1/sqrt2`

When `cosx=1/sqrt2=cos(pi/4)`

`=>x=2npi+-pi/4" where "n inZZ`

When `cosx=-1/sqrt2=-cos(pi/4)=cos((3pi)/4)`

`=>x=2npi+-(3pi)/4" where "n inZZ`