求这道题的解? 20
1个回答
展开全部
f(x)=向量a.向量b=2(cosx)^2-1-sin(2x+π/6)=cos2x-【√3/2*sin2x+1/2*cos2x】
=-√3/2*sin2x+1/2*cos2x=-sin(2x-π/3),
则其减区间满足2kπ-π/2<2x-π/3<2kπ+π/2,
得kπ-π/12<x<kπ+5π/12,
又x∈(0,π),即(0,5π/12)和(11π/12,π),
因为x∈(0,π),得-π/3<2x-π/3<5π/3,
画草图,k=-3f(x)恰有两个交点,
则k≠±3或√3/2
=-√3/2*sin2x+1/2*cos2x=-sin(2x-π/3),
则其减区间满足2kπ-π/2<2x-π/3<2kπ+π/2,
得kπ-π/12<x<kπ+5π/12,
又x∈(0,π),即(0,5π/12)和(11π/12,π),
因为x∈(0,π),得-π/3<2x-π/3<5π/3,
画草图,k=-3f(x)恰有两个交点,
则k≠±3或√3/2
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询