已知向量a=(5√3cosx,cosx),b=(sinx,2cosx),设函数f(x)=a*b+|b|^2+3/2

已知向量a=(5√3cosx,cosx),b=(sinx,2cosx),设函数f(x)=a*b+|b|^2+3/2
若f(θ/2)=9 求f(θ+ π/3)的值
求使f(x)≥5成立的x的取值集合
feefee哭 1年前 已收到2个回答 举报

qq_飞鱼 幼苗

共回答了14个问题采纳率:92.9% 举报

f(x)=5√3sinxcosx+2cos^2x+sin^2x+4cos^2x+3/2
=5√3/2sin2x+6cos^2+sin^2x+3/2
=5√3/2sin2x+6cos^2x-3+3+sin^2x-1/2+1/2+3/2
=5√3/2sin2x+3cos2x-1/2cos2x+5
=5√3/2sin2x+5/2cos2x+5
=5sin(2x+π/6)+5
(1)
2x+π/6=π/2+2kπ(k∈Z)
x=π/6+kπ
2x+π/6=3π/2+2kπ(k∈Z)
x=2π/3+kπ
f(x)在[π/6+kπ,2π/3+kπ]单调递减
f(x)的值域为[5/2,10]
(2)
x∈[π/6,π/2]
2x+π/6∈[π/2,7π/6]
5sin(2x+π/6)+5=8
sin(2x+π/6)=3/5
cos(2x+π/6)=-4/5
sin[2(x-π/12)+π/6]
=sin[2x-π/6+π/6]
=sin(2x+π/6)cosπ/6-cos(2x+π/6)sinπ/6
=3/5*√3/2-(-4/5)*1/2
=3√3/10+4/10
=(3√3+4)/10

1年前

5

ccf_jcc 幼苗

共回答了1576个问题 举报

答:
向量a=(5√3cosx,cosx),b=(sinx,2cosx),设函数f(x)=a*b+|b|^2+3/2
f(x)=5√3cosxsinx+cosx*2cosx+(sinx)^2+(2cosx)^2+3/2
=(5√3/2)sin2x+5(cosx)^2+1+3/2
=(5√3/2)sin2x+(5/2)(cos2x+1)+5/2
=(5√3/2...

1年前

2
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