f(x)=(√3)[sin(x/2)][cos(x/2)]+cos²(x/2)?
是嗎?
如果是的話：
最小正周期：2π/(1/2)=4π
單調(diào)遞增區(qū)間是：x∈(4kπ-7π/6,4kπ+5π/6),其中：k∈N
f(x)=(√3)[sin(x/2)][cos(x/2)]+cos²(x/2)
f(x)=cos(x/2)[(√3)sin(x/2)+cos(x/2)]
f(x)=2cos(x/2){[(√3)/2]sin(x/2)+(1/2)cos(x/2)}
f(x)=2cos(x/2)[cos(π/6)sin(x/2)+sin(π/6)cos(x/2)]
f(x)=2cos(x/2)sin(x/2+π/6)
f(x)=sin[(x/2+x/2+π/6)/2]cos[(x/2+π/6-x/2)/2]
f(x)=sin(x/2+π/12)cos(π/12)
最小正周期：2π/(1/2)=4π
f‘(x)=(1/2)cos(x/2+π/12)cos(π/12)
令：f'(x)＞0,即：(1/2)cos(x/2+π/12)cos(π/12)＞0
整理,有：cos(x/2+π/12)＞0
得：2kπ-π/2＜x/2+π/12＜2kπ+π/2,其中：k∈N
即：4kπ-7π/6＜x＜4kπ+5π/6
f(x)的單調(diào)遞增區(qū)間是：x∈(4kπ-7π/6,4kπ+5π/6),其中：k∈N