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
由tanx=tan(x+π)
y=tan(2x+π/3)=tan(2x+π/3+π)=tan(2(x+π/2)+π/3)
所以函数y=tan(2x+π/3),x≠5π/12+kπ/2(k∈z)的周期为π/2
答
与X的值无关，主要看X前面的系数tan周期是π，所以该函数的周期为T=π/2
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