求三角函数周期、最值问题f(x)=sin(2x+3/π)+sin(2x-3/π)+2cos²x-1,x∈R(1

解析：
f(x)=sin(2x+ 3分之π)+sin(2x- 3分之π)+2cos²x-1
=sin2x*cos(3分之π) +cos2x*sin(3分之π)+sin2x*cos(3分之π) -cos2x*sin(3分之π)+cos2x
=2sin2x*cos(3分之π) +cos2x
=sin2x + cos2x
=√2*(sin2x*√2/2 + cos2x*√2/2)
=√2*sin(2x+ π/4)
则可知函数f(x)的最小正周期T=2π/2=π
若x∈[-π/4,π/4],即2x∈[-π/2,π/2],那么：2x+ π/4∈[-π/4,3π/4]
所以当2x+ π/4=π/2即x=π/8时,函数f(x)有最大值为√2；
当2x+ π/4=-π/4即x=-π/4时,函数f(x)有最小值为√2*(-√2/2)=-1.