y=sin(x+π/3)+2定义域怎么求?

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y=sin(x+π/3)+2定义域怎么求?y=sin(x+π/3)+2定义域怎么求?y=sin(x+π/3)+2定义域怎么求?y∈[1,3]当y=1时,sin(x+π/3)=-1,x+π/3=2kπ-

y=sin(x+π/3)+2定义域怎么求?
y=sin(x+π/3)+2定义域怎么求?

y=sin(x+π/3)+2定义域怎么求?
y∈[1,3]
当y=1时,sin(x+π/3)=-1,x+π/3=2kπ-π/2,x=kπ-5π/12,k∈Z
当y=3时,sin(x+π/3)=1,x+π/3=2kπ+π/2,x=kπ+π/12,k∈Z
∴当x∈[kπ-5π/12,kπ+π/12]时,y∈[1,3]

y=sin(x)的定义域为R,X加上一个常数其定义域不变,所以定义域仍是R