Question

The general solution of the trigonometric equation `sinx+cosx=1` is given by `x=2npi,n0,+-1,+-2` `x=2npi+pi/2; n=0,+-1,+-2, dot` `x=npi+(-1)^npi/4-pi/4n=0,+-1,+-2, ` `non eoft h e s e`

Answer 1

`sinx+cosx=1`
`sinx=1-cosx`
`1-sinx=sqrt(1-sin^2x)`
`(1-sinx)^2=(1-sin^2x)`
`1+sin^2x-2sinx=1-sin^2x`
`2sin^2x-2sinx=0`
`2sinx(sinx-1)=0`
`sinx=0`
`x=pmnpi`
`n in Z`
`sinx-1=0`
`sinx=1`
`x=pi/2+2npi`
`n in Z`.