y=sin(2x+π/6)+cos²x-1的周期、单调区间、最值、对称轴
问题描述：
y=sin(2x+π/6)+cos²x-1的周期、单调区间、最值、对称轴
答
y=sin(2x+π/6)+cos²x-1=sin2xcos(π/6)+cos2xsin(π/6)+(1/2)cos2x-(1/2)=(√3/2)sin2x+cos2x-(1/2)=(√7/2)*[(√3/√7)sin2x+(2/√7)cos2x]-(1/2)=(√7/2)sin(2x+φ)-(1/2)……0