y=sin(2x+π/6)+cos²x-1的周期、单调区间、最值、对称轴

f(x)=2sinxcos(π/6)+cosx
=√3sinx+cosx
=2sin(x+π/6)
周期为2π
2kπ+π/2≤x+π/6≤2kπ+3π/2,k∈Z.
解得2kπ+π/3≤x≤2kπ+4π/3
在[0,2π]上的单调递减区间是[π/3,4π/3]

y=sin(x+π/6)+cosx
=sinxcosπ/6+cosxsinπ/6+cosx
=sinx*√3/2+cosx*1/2+cosx
=sinx*√3/2+cosx*3/2
=√3(sinx*1/2+cosx*√3/2)
=√3(sinx*cosπ/3+cosx*sinπ/3)
=√3sin(x+π/3)
所以最小正周期是T=2π

f(x)=cos²x-(√3)cosxsinx+1
f(x)=2[cos²x-(√3)cosxsinx+1]/2
f(x)=[2cos²x-2(√3)cosxsinx+2]/2
f(x)=[2cos²x-1-(√3)·2cosxsinx+3]/2
f(x)=[cos(2x)-(√3)sin(2x)+3]/2
f(x)=(1/2)cos(2x)-[(√3)/2]sin(2x)+3/2
f(x)=sin(π/6)cos(2x)-cos(π/6)sin(2x)+3/2
f(x)=sin(π/6-2x)+3/2
f(x)=3/2-sin(2x-π/6)
f(x)的单调递增区间是：2kπ-π/2≤2x-π/6≤2kπ+π/2,其中：k=0、±1、±2、±3……,下同.
2kπ-π/3≤2x≤2kπ+2π/3
kπ-π/6≤x≤kπ+π/3
即,所求单调递增区间是：x∈[kπ-π/6,kπ+π/3]

有了上面求出的f(x)=3/2-sin(2x-π/6),第二问就很简单了.
就留给楼主做练习吧

(1) f(0)=-1/2+1=1/2
(2) f(x)=sin(x)*sqrt(3)/2-cos(x)/2+cos(x)=sin(x)*sqrt(3)/2+cos(x)/2=sin(x+pi/6)
最大值为1,最小值sqrt(3)/2
(3) f(a+π/3)=sin(a)*cos(pi/6+pi/3)+cos(a)*sin(pi/6+pi/3)=1/3

接触方程即可求出