已知向量a=(sinx,2√ 3sinx),b=(mcosx,-sinx),定义f(x)=a*b+√ 3,且x=π/6是函数Y=F(X)的零点

(1)
f(x)=msinxcosx-2√3sin^2(x)+√3 ,因为x=π/6是函数的零点,所以
0=m(√3/4)-√3/2+√3 ==>m= - 2
f(x)= - sin2x-√3(1-cos2x)+√3
=2cos(2x+π/6)
由 -π+2kπ≤2x+π/6≤2kπ得单调增区间是：【-7π/12+kπ,-π/12+kπ】
由 2kπ≤2x+π/6≤π+2kπ得单调增区间是：【-π/12+kπ,5π/12+kπ】
(2)
f(x+θ)=2cos(2x+2θ+π/6)是奇函数,所以当x=0时,上式为零
0=2cos(2θ+π/6)=0 ==>θ=π/6
(3)
2cos(2A+π/6)= - 1 ==>cos(2A+π/6)= - 1/2
A=π/4 ,则正弦定理得：1/sinπ/4=√2/sinB ==>sinB=1
B=π/2 ==>C=π/4