求函数y=-tan(2x+π/3),x=5π/12+kπ/2,(k∈z)的周期

tan周期是π这里x系数是2所以T=π/2x=5π/12+kπ/2是不是求值?y=-tan(2x+π/3)=-tan(5π/6+kπ+π/3)=-tan(5π/6+π/3)=-(tan5π/6+tanπ/3)/(1-tan5π/6*tanπ/3)=-(-√3/3+√3)/(1+1)=-√3/3