已知函数f(x)=根号3sinxcosx+cos²x+1/2,x属于R,求f(x)的递减区间

f(x)=√3sinxcosx+cos²x+1/2=√3/2 sin2x+1/2(2cos²x-1）+1=√3/2 sin2x+1/2cos2x+1=cosπ/6sin2x+sinπ/6cos2x+1=sin（π/6+2x）+1f(x)的递减区间:2kπ+π/2≤π/6+2x≤2kπ+3π/22kπ+π/3≤2x≤2kπ+4...