Question

Question #aa66a

Answer 1

Write this as follows

`sinx-cosx=1`

Multiply with `sqrt2/2` both sides

`sqrt2/2*(sinx-cosx)=sqrt2/2`

`sin(x-pi/4)=sin(pi/4)`

Hence the general solutions are

`x-pi/4=2*κ*pi+pi/4` and `x-pi/4=2*k*pi+3pi/4`

or

`χ=2*k*pi+pi/2` and `x=2*k*pi+pi`

Because x belongs in `(0,2*pi]` acceptable solutions are

`x=pi` and `x=pi/2`