举报
vip_111
(x)=sin(2x+π/6)+sin(2x-π/6)-cos2x+1 =(sin2xcosπ/6+cos2xsinπ/6)+(sin2xcosπ/6-cos2xsinπ/6)-cos2x+1 =2(sin2x)cosπ/6-2cos2xsinπ/6+1 =2sin(2x-π/6)+1 最小正周期T=2π/2=π sin(2x-π/6)=1或-1,即2x-π/6=kπ+π/2,即x=kπ/2+π/3是对称轴,【k是整数,下同】 sin(2x-π/6)=0,即2x-π/6=kπ,即x=kπ/2+π/12,点P(kπ/2+π/12,1)是对称中心 sin(2x-π/6)=1,即2x-π/6=2kπ+π/2,x=kπ+π/3,T=π,单调递增区间为[kπ-2π/3,kπ+π/3] 函数的最大值3,最小值-1,即值域[-1,3]