解:
(1)
即sin(π/3+x)≠0
π/3+x≠π+2kπ
x≠2π/3+ 2kπ k∈Z
(2)
cosa=3/5
则
sina=4/5
f(a)=cos(a+π/6)/sin(a+π/3)
=[cosacosπ/6-sinasinπ/6]/[sinacosπ/3+cosasinπ/3]
=[(3/5)(根号3/2)-(4/5)(1/2)]/[(4/5)(1/2)+(3/5)(根号3/2)]
=[(3根号3 -4)/10]/[(4+3根号3)/10]
=(3根号3 -4)/(3根号3 +4)
=(3根号3 -4)^2 /11