f(x)=2-cos(2x-π/3)-2sin²x,x∈{0,π/2},求f(x)值域

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f(x)=2-cos(2x-π/3)-2sin²x,x∈{0,π/2},求f(x)值域
f(x)=2-cos(2x-π/3)-2sin²x,x∈{0,π/2},求f(x)值域

f(x)=2-cos(2x-π/3)-2sin²x,x∈{0,π/2},求f(x)值域
f(x)=2-cos(2x-π/3)-2sin²x
=2-1/2cos2x-√3/2sin2x-1+cos2x
=1+1/2cos2x-√3/2sin2x
=1+cos(2x+π/3)
因为x∈{0,π/2},则2x+π/3∈{π/3,4π/3},
所以f(x)∈{0,3/2}即:f(x)的值域为:{0,3/2}

1-根号3 ≤ f(x) <1/2