已知f(x)=sin^2x+2sinxcosx+3cos^2x,x∈(0,兀),求函數(shù)的最小正周期和值域;求函數(shù)的單調(diào)遞增區(qū)間
已知f(x)=sin^2x+2sinxcosx+3cos^2x,x∈(0,兀),求函數(shù)的最小正周期和值域;求函數(shù)的單調(diào)遞增區(qū)間
數(shù)學(xué)人氣:321 ℃時間:2019-09-29 03:07:54
優(yōu)質(zhì)解答
2cos²x-1=cos2xf(x)=sin²x+2sinxcosx+3cos²x=(sin²x+cos²x)+(2cos²x-1)+1+2sinxcosx=2+sin2x+cos2x=2+(√2)sin[2x+(π/4)]T=2π/ω=πx∈(0,π)[2x+(π/4)]∈(π/4,9π/4)f(x)max=2+(...
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