设x∈[0,2π]且√1-sin2x=sin x-cos x求x的取值范围

证明：要证2sin2y-sin2x=sin(2x+2y),
即要证2sin2y=sin2x+sin(2x+2y)=2sin(2x+y)cosy（利用和差化积）
即要证4sinycosy=2sin(2x+y)cosy
两边消去2cosy,得2siny=sin(2x+y)
也即要证2siny=sin2xcosy+sinycos2x,即siny(2-cos2x)=sin2xcosy,即tany=sin2x / (2-cos2x),
由万能公式知sin2x=2tanx / (1+tan^2 x),cos2x=(1-tan^2 x) / (1+tan^2 x)
所以tany=sin2x / (2-cos2x)又可化为tany=2tanx / (1+3tan^2 x),只需证明此式成立,原式即得证
而由已知条件tan(x+y)=3tanx 即 (tanx+tany) / (1-tanxtany)=3tanx
即tanx+tany=3tanx-3tan^2 xtany
即(1+3tan^2 x)tany=2tanx
也就是tany=2tanx / (1+3tan^2 x)
所以,原式得证.
注：平方符号不知如何打,用tan^2 x表示tanx的平方