Question

已知函数y=2根号3 sinxcosx+cos2x（x∈R）,问1 函数的最小正周期,并求对称轴方程2 单调递增区间【详

Answer 1

y=2根号3 sinxcosx+cos2x=2sin(2x+π/6) 最小正周期T=π

单调递增区间2kπ-π/2≤2x+π/6≤2kπ+π/2 ,

2kπ-2π/3≤2x≤2kπ+π/3

kπ-π/3≤x≤kπ+π/6

y=2sinx(x∈R) 图象纵坐标不变,横坐标变为原来的1/2倍得函数y=2sin2x

再将y=2sin2x图象向左平移|π/12|个单位得y=2sin(2x+π/6)图象