Question

已知0＜x＜2π，若根号下(1-Sin2x)=Sinx-Cosx，则x的取值范围是？（要详细步骤哦）

Answer 1

答案:x=π/4,x=5π/4,x=0,x=π/2
由(1-sin2x)=sinx-cosx得
(sinx-cosx)^2=sinx-cosx
(sinx-cosx)(sinx+cosx-1)=0,故得
sinx-cosx=0或sinx+cosx-1=0
由sinx-cosx=0得
tanx=1,x=π/4,x=5π/4,
由sinx+cosx-1=0得
√2sin(x+π/4)=1
sin(x+π/4)=√2/2
x+π/4=π/4或x+π/4=3π/4
x=0或x=π/2

Answer 2

1-sin2x=(sinx-cosc)^2
即sinx-cosx>=0
45<=x<=225

Answer 3

根号下(1-Sin2x)=sinx-cosx>=0,所以解的π/4<=x<=5π/4.