已知函数f(x)=√3sinxcosx cos눀x a的最小正周期和单调递减区间

2025-05-14 07:02:15
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回答(1):

f(x)=√3sinxcosx +cos²x+ a=√3/2 sin2x+(cos2x+1)/2+a
=√3/2 sin2x+cos2x/2+1/2+a=sin(2x+π/6)+a
T=2π/2=π
2kπ+π/2<=x+π/6<=2kπ+3π/2 2kπ+π/3<= x <=2kπ+4π/3