求函数y=-tan(2x+π/3),x=5π/12+kπ/2,(k∈z)的周期
问题描述：
求函数y=-tan(2x+π/3),x=5π/12+kπ/2,(k∈z)的周期
答
tan周期是π
这里x系数是2
所以T=π/2
x=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