y=sin2x-2(1+cos2x)/2+3
=sin2x-cos2x+1
=√2(√2/2*sin2x+√2/2zos2x)+1
=√2(sin2xzosπ/4+zos2xsinπ/4)+1
=√2sin(2x+π/4)+1
所以最大值√2+1,最小值√2-1
sin增区间(2kπ-π/2,2kπ+π/2)
2kπ-π/2
y=sin2x-2(1+cos2x)/2+3
=sin2x-cos2x+1
=√2(√2/2*sin2x+√2/2zos2x)+1
=√2(sin2xzosπ/4+zos2xsinπ/4)+1
=√2sin(2x+π/4)+1
所以最大值√2+1,最小值√2-1
sin增区间(2kπ-π/2,2kπ+π/2)
2kπ-π/2