已知函数f(x)=根号3倍sinx/2cosx/2+cos^2x/2+1/2求f(x)单调减区间
问题描述：
已知函数f(x)=根号3倍sinx/2cosx/2+cos^2x/2+1/2求f(x)单调减区间
答
f(x)=根号3倍sinx/2cosx/2+cos^2x/2+1/2
=√3/2*sinx+(cosx+1)/2+1/2
=sin(x+π/6)+3/2
所以f(x)单调减区间为[2nπ+π/2-π/6,2nπ+3π/2-π/6]
既[2nπ+π/3,2nπ+4π/3]
答
f(x)=根号3倍sinx/2cosx/2+cos^2x/2+1/2
=(√3/2)sinx+1/2*【2*cos^2x/2-1】+1
=(√3/2)sinx+1/2cosx+1
=sin(x+Π/6)+1
就可以知道f(x)的单调减区间是【Π/3+2kΠ,4Π/3+2kΠ】
答
f(x)=√3sin(x/2)cos(x/2)+cos²(x/2)+1/2
=√3/2sinx+(cosx+1)/2+1/2
=√3/2sinx+1/2cosx+1
=sin(x+π/6)+1
令2kπ+π/2≤x+π/6≤2kπ+3π/2 (k∈N)
则2kπ+π/3≤x≤2kπ+4π/3 (k∈N)
∴f(x)的单调递减区间为[2kπ+π/3,2kπ+4π/3](k∈N)