Question

做出函数y=1/2+cosx,x=∈[-π,π]的大致图像,并分别写出使y>0和y≤1的x的取值范围.

Answer 1

y=1/2+cosx,x=∈[-π,π]

x -π -π/2 0 π/2 π

cosx -1 0 1 0 -1

y -1/2 1/2 3/21/2 -1/2

由图得到

y>0时,x∈[-π,π]时,

x取值范围 [-2π/3, 2π/3],

x∈R时,x取值范围

[2kπ-2π/3, 2kπ+2π/3],k∈Z

y≤1,

x∈[-π,π]时,x取值范围x∈[-π,-π/3]U[π/3,π],

x∈R时,x取值范围 [2kπ+π/3,2kπ+5π/3]k∈Z