2sinπ/6x=1-cosπ/3x
f（x）=2sinπ/6x+sin（π/3x+π/6）—1
=（1-cosπ/3x)+sinπ/3xcosπ/6+cosπ/3xsinπ/6-1
=√3/2sinπx/3-1/2cosπx/3
=sin(πx/3-π/6)
∴f(x)max=1,f(x)min=-1
∴f(x)的值域?yàn)閇-1,1]希望能幫到你，不懂可以問！
已知函數(shù)fx＝2sinπ/6x+sin（π/3x+π/6）-1 （1）求f（x）的值域
已知函數(shù)fx＝2sinπ/6x+sin（π/3x+π/6）-1 （1）求f（x）的值域
數(shù)學(xué)人氣：718 ℃時(shí)間：2020-03-23 06:33:28
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