Giải pt :$\sqrt[]{1-sinx}+\sqrt[]{1-cosx} = 1$

`\sqrt{1-sinx}+\sqrt{1-cosx}=1`

`<=>1-sinx+2\sqrt{(1-sinx)(1-cosx)}+1-cosx=1`

`<=>2\sqrt{1-(sinx+cosx)+sinxcosx}=sinx+cosx-1`

Đặt `sinx+cosx=t(tin[-\sqrt{2}; \sqrt{2}])` ta được:

`2\sqrt{1-t+(t^2-1)/2}=t-1`

`<=>4-4t+2t^2-2=t^2-2t+1`

`<=>t^2-2t+1=0`

`<=>(t-1)^2=0`

`<=>t-1=0`

`<=>t=1(TM)`

`<=>\sqrt{2}sin(x+\pi/4)=1`

`<=>sin(x+\pi/4)=1/\sqrt{2}`

`<=>[(x+\pi/4=\pi/4+k2\pi),(x+\pi/4=\pi-\pi/4+k2\pi):}`

`<=>[(x=k2\pi),(x=\pi/2+k2\pi):}`

Vậy `S={k2\pi; \pi/2+k2\pi|kinZZ}`