Question

How do you solve `sqrt2sin2x=1` between the interval `0<=x<=2pi`?

Answer 1

We have that

`sqrt2 sin2x=1=>sin2x=1/sqrt2=>sin2x=sqrt2/2=> sin2x=sin(pi/4)`

Hence from the last relation we get

`2x=2kpi+pi/4=>x=kpi+pi/8`

or

`2x=2kpi+pi-pi/4=>2x=2kpi+3pi/4=>x=kpi+3pi/8`

Because `0<=x<=2pi` acceptable solutions are

`x_1=pi/8,x_2=9pi/8,x_3=3pi/8,x_4=11pi/8`