Question

How do you solve `2sin^2theta=1`?

Answer 1

`theta=((2k+1))/4pi , k in mathbbZ`

Explanation:

`2sin^2theta=1`
`sin^2 theta=1/2`
`sintheta=sqrt (1/2) or sintheta=-sqrt(1/2)`
`sintheta=sqrt2/2 or sintheta=-sqrt2/2`
`theta=pi/4+2kpi or theta=pi-pi/4+2kpi or theta=-pi/4+2kpi or theta=-(pi-pi/4)+2kpi, k in mathbbZ`
`theta=pi/4+kpi or theta=-pi/4+kpi, k in mathbbZ`
`theta=pi/4+kpi/2, k in mathbbZ`
or written a bit differently
`theta=((2k+1))/4pi , k in mathbbZ`