若f(x)=asin(kx+π/3)和g(x)=btan(kx-π/3)(k>0),若它们的最小正周期之和为3π/2,

最小正周期之和为3π/2,2π/k+π/k=3π/2k=2f(π/2)=asin(π+π/3)=-a√3/2g(π/2)=btan(π-π/3)=-b√3所以,a=2bf(π/4)=asin(π/2+π/3)=a/2g(π/4)=btan(π/2-π/3)=b√3/3所以,a/2=-b+1解方程组得：a=1,b=1/2所...