f(x)=sin2x+(1-2sin²x)-1
=sin2x+cos2x-1
=√2(√2/2*sin2x+√2/2cos2x)-1
=√2(sin2xcosπ/4+cos2xsinπ/4)-1
=√2sin(2x+π/4)-1
T=2π/2=π
-1<=sin(2x+π/4)<=1
所以sin(2x+π/4)=1时最大
所以2x+π/4=2kπ+π/2
x=kπ+π/8
所以x∈{x|x=kπ+π/8},f(x)最大值=√2-1