已知函数y=根号2cos平方x-根号2sin平方x+2根号2sinxcosx+1,x属于R,（1）求函数的值域和最小正周期.（2）求该函数去最值时自变量x的取值范围

解
y=√2cos²x-√2sin²x+2√2sinxcosx+1
=√2(cos²x-sin²x)+√2sin2x+1
=√2sin2x+√2cos2x+1
=2(√2/2sin2x+√2/2cos2x)+1
=2(sin2xcosπ/4+cos2xsinπ/4)+1
=2sin(2x+π/4)+1
∵sin(2x+π/4)∈[-1,1]
∴-1