f(x)=2sin(2x+2A+π/6)+1,A为常数,（1）设∠A,∠B是三角形ABC的内角,若f(B)=0,求∠C

f(B)=2sin(2B+2A+π/6)+1=0
sin(2B+2A+π/6)=-1/2=sin7π/6或sin11π/6
2A+2B+π/6=7π/6或11π/6
A+B=π/2或5π/6
C=π-(A+B)=π/2或π/6

∠A,∠B是三角形ABC的内角,所以sin(A+B)=sinC,cos(A+B)=-cosC
f(B)=2sin(2B+2A+π/6)+1
=√3sin2(A+B)+cos2(A+B)+1
=2√3sin(A+B)cos(A+B)+2[cos(A+B)]^2
=-2√3sinC*cosC+2[cosC]^2
=2cosC*（cosC-√3sinC）
=0
所以cosC=0或tanC=√3/3
C=π/2或C=π/6

f(x)=2sin(2x+2A+π/6)+1=2[sin2(x+A）cos(π/6)+cos2(x+A)sin(π/6)]+1=√3sin2(x+A）+cos2(x+A)+1
设∠A,∠B是三角形ABC的内角，若f(B)==√3sin2(B+A）+cos2(B+A)+1=√3sin2(π-C）+cos2(π-C)+1=0
即f(B)=2sin[(π/6-2C)+1=2sin[(π/...