函数f(x)=[2sin(x+π/3)+sinx]cosx-根3sin^2x,(x∈R).
问题描述：
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)
数学人气：760 ℃时间：2019-08-20 04:04:28
优质解答
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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答
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,+∞）.