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已知函数f(x)=2√3sinxcosx-2cosx+1.求函数的最小正周期?若a∈(0,二分之派),且f(a)=1,求a的值?
人气:332 ℃ 时间:2020-04-16 03:38:01
解答
f(x)=2√3sinxcosx-2cos^2 x+1.=√3sin2x -cos2x=2(√3/2sin2x -1/2 cos2x)=2(sin2xcosπ/6-cos2xsinπ/6)=2sin(2x-π/6)∴最小正周期T=2π/2=πf(a)=2sin(2a-π/6)=1sin(2a-π/6)=1/22a-π/6=2kπ+π/6 或2a-π/6...
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