已知函数f(x)=[2sin(x=π/3)+sinx]cosx-√3sin²x,x∈R.若存在x∈[0,5π/
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已知函数f(x)=[2sin(x=π/3)+sinx]cosx-√3sin²x,x∈R.若存在x∈[0,5π/12],使不等式f(x)<m成立,求实数m的取值范围
![已知函数f(x)=[2sin(x=π/3)+sinx]cosx-√3sin²x,x∈R.若存在x∈[0,5π/](/uploads/image/z/16073000-8-0.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3D%5B2sin%28x%3D%CF%80%2F3%29%2Bsinx%5Dcosx-%E2%88%9A3sin%26%23178%3Bx%2Cx%E2%88%88R.%E8%8B%A5%E5%AD%98%E5%9C%A8x%E2%88%88%5B0%2C5%CF%80%2F)
f(x)=[2sin(x+π/3)+sinx]cosx-√3sin²x
=[2sinxcosπ/3+2cos2xsinπ/3+sinx]cosx-√3sin²x
=2sinxcosx+√3(cos²x-sin²x)
=sin2x+√3cos2x
=2(1/2*sin2x+√3/2*cos2x)
=2sin(2x+π/3)
∵存在x∈[0,5π/12],使不等式f(x)<m成立
∴需f(x)min-1
再问: 使不等式f(x)<m成立是需要f(x)max
=[2sinxcosπ/3+2cos2xsinπ/3+sinx]cosx-√3sin²x
=2sinxcosx+√3(cos²x-sin²x)
=sin2x+√3cos2x
=2(1/2*sin2x+√3/2*cos2x)
=2sin(2x+π/3)
∵存在x∈[0,5π/12],使不等式f(x)<m成立
∴需f(x)min-1
再问: 使不等式f(x)<m成立是需要f(x)max
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