一道数学会考题：已知函数f(x)=sin(2x+π/3)+cos(2x-π/6)

（1）将函数f(x)的解析式化为f(x)=Asin(ωx+φ) 的形式，并指出它的最小正周期
（2）当x∈[- π/6,π/6]时，求f(x)的最小值和相应x的值

1)f(x)=sin(2x+π/3)+sin(π/2-2x+π/6)
=sin(2x+π/3)+sin(-2x+2π/3)
=sin(2x+π/3)+sin(2x+π/3)
=2sin(2x+π/3)
T=2π/2=π

2)-π/6= 0=<2x+π/3<=2π/3
最小值为2x+π/3=0时，f(x)=0
此时x=-π/6

(1)f(x)=sin2xcosπ/3+sinπ/3cos2x+cos2xcosπ/6+sin2xsinπ/6=sin2x+根号3cos2x=2sin(2x+π/3) T=π (2)当x∈[- π/6,π/6]时,2x+π/3∈[ 0,2π/3]所以当x=-π/6有最小值为2sin0=0

f(x)=sin(2x+π/6)-cos(2x+π/3)+2cos²x =根号下3sin2x/2+cos2x/2-cos2x/2+根号下3sin2x/2+2cos²x-1+1 =根号下3sin2x+cos2x+1 =2(根号下3sin2x/2+cos2x/2)+1 =2sin(2x+π/6)+1 f(π/12)=2sin(π/3)+1=根号下3+1

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