因?yàn)楫?dāng)(nx/6+π/5)=π/2+2πk,f(x)=max=1
so x=12kπ/n+3π/n-6π/5n
x=(60kπ+9π)/5n f(x)=max
因?yàn)楫?dāng)(nx/6+π/5)=-π/2+2πk,f(x)=min=-1
so x=12kπ/n-3π/n-6π/5n
f(x)=min
T=2π/(n/6)=12π/n
因?yàn)橐沟脙蓴?shù)之間至少有一個(gè)最大值和最小值
要使T