函数f(x)=[2sin(x+π/3)+sinx]cosx-根3sin^2x,(x∈R).1)求函数f(x)的最小正周期2)若存在x0∈[0,5π/12],使不等式f(x0)
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![函数f(x)=[2sin(x+π/3)+sinx]cosx-根3sin^2x,(x∈R).1)求函数f(x)的最小正周期2)若存在x0∈[0,5π/12],使不等式f(x0)](/uploads/image/z/616370-50-0.jpg?t=%E5%87%BD%E6%95%B0f%28x%29%3D%5B2sin%28x%2B%CF%80%2F3%29%2Bsinx%5Dcosx-%E6%A0%B93sin%5E2x%2C%28x%E2%88%88R%29.1%29%E6%B1%82%E5%87%BD%E6%95%B0f%28x%29%E7%9A%84%E6%9C%80%E5%B0%8F%E6%AD%A3%E5%91%A8%E6%9C%9F2%EF%BC%89%E8%8B%A5%E5%AD%98%E5%9C%A8x0%E2%88%88%5B0%2C5%CF%80%2F12%5D%2C%E4%BD%BF%E4%B8%8D%E7%AD%89%E5%BC%8Ff%28x0%29)
函数f(x)=[2sin(x+π/3)+sinx]cosx-根3sin^2x,(x∈R).1)求函数f(x)的最小正周期2)若存在x0∈[0,5π/12],使不等式f(x0) 函数f(x)=[2sin(x+π/3)+sinx]cosx-根3sin^2x,(x∈R).1)求函数f(x)的最小正周期2)若存在x0∈[0,5π/12],使不等式f(x0)
函数f(x)=[2sin(x+π/3)+sinx]cosx-根3sin^2x,(x∈R).
1)求函数f(x)的最小正周期
2)若存在x0∈[0,5π/12],使不等式f(x0)
f(x)=[2sin(x+π/3)+sinx]cosx-根3sin^2x
=[sinx+(√3)cosx+sinx]cosx-(√3)(sinx)^2
=2sinxcosx+(√3)[(cosx)^2-(sinx)^2]
=sin2x+(√3)cos2x
=2sin(2x+π/3),(x∈R).
1)函数f(x)的最小正周期是π.
2)x0∈[0,5π/12],
∴2x0+π/3∈[π/3,7π/6],
∴f(x0)的最大值是2,
∴m的取值范围是(2,+∞).
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