已知函数fx=根号2sin(2x+3分之π)的单调性和最小正周期，还有在[0,,2分之π】的值域

详细步骤！！！！因为很急！所以谢谢！

f(x)=√2sin(2x+π/3)
=√2sin(2x+2π+π/3)
=√2sin[2(x+π)+π/3]
∴最小正周期=π
∵2kπ-π/2<2x+π/3<2kπ+π/2
即 kπ-5π/12<x<kπ+π/12
∴f(x)在kπ-5π/12<x<kπ+π/12单增
∵2kπ+π/2<2x+π/3<2kπ+3π/2
即 kπ+π/12<x<kπ+7π/12
∴f(x)在kπ+π/12<x<kπ+7π/12单减
∵0<x<π/2
即 2π/3<2x+π/3<5π/3
f(x)在2x+π/3∈（2π/3，3π/2)递减，在2x+π/3∈（3π/2，5π/3）递增，
在2x+π/3=2π/3 函数值=√6/2，最大
在2x+π/3=3π/2时，函数值最小=-√2
∴函数在[0,,2分之π】的值域：[-√2, √6/2]

2x。=2π，x。=π 最小正周期: π 主值单调递增区间： -π/2<=2x+3分之π<=π/2 -5π/6<=2x<=π/6 -5π/12<=x<=π/12 主值单调递减区间： 。。。 在[0,,2分之π】区间 3分之π<=π+3分之π 则 -根号3<=f(x)<=2

f（x）= sin^2(2x+π/4) + √3 cos^2(2x) = {sin2xcosπ/4+cos2xsinπ/4}^2 + √3 cos^2(2x) = 1/2(sin2x+cos2x)^2 + √3 cos^2(2x) = 1/2(1+2sin2xcos2x) + √3 * (cos4x+1)/2 = 1/2 + 1/2sin4x +√3/2 cos4x + √3/2 = sin4xcosπ/3 + cos4xsinπ/3 + (1+√3)/2 = sin(4x+π/3) + (1+√3)/2 最小正周期=2π/4=π/2 当4x+π/3∈（2kπ+π/2，2kπ+3π/2)，其中k∈z时单调减 单调减区间：x∈（kπ/2+π/24，kπ/2+7π/24)，其中k∈z 若x∈【-π/6，π/6】 4x+π/3∈（-π/3，π) 4x+π/3=-π/3，x=-π/6时，函数有最小值f(x)min=sin(-π/3) + (1+√3)/2 = -√3/2+(1+√3)/2 = 1/2 4x+π/3=π/2，x=π/24时，函数有最大值f(x)max=sin(π/2) + (1+√3)/2 = 1+(1+√3)/2 =(3+√3)/2