已知f(x)=2sin(x+θ/2)cos(x+θ/2)+2倍根號3倍的cos(x+θ/2)^2-根號31,化簡f(x)的解析式2若0≤θ≤派,問θ為多少時,可使函數f(x)為偶函數?3,在2成立的條件下,求滿足f(x)=1,x∈[-派,派]的x的集合

熱心網友

1.f(x)=2sin(x+θ/2)cos(x+θ/2)+2*根號(3)*cos(x+θ/2)^2-根號(3)==4cos(x+θ/2)[(1/2)sin(x+θ/2)+(根號(3)/2)*cos(x+θ/2)]-根號(3)==4cos(x+θ/2)sin(x+θ/2+π/3)-根號(3)=2[sin(2x+θ+π/3)+sin(π/3)]-根號(3)==2sin(2x+θ+π/3)2.f(x)為偶函數,f(x)=f(-x),sin(2x+θ+π/3)=sin(-2x+θ+π/3),2x+θ+π/3-2x+θ+π/3=π+2kπ,θ=π/6+kπ,k=0,θ=π/63.f(x)=2sin(2x+π/6+π/3)=2cos(2x)=12x=π/3+2kπ,-π/3+2kπ,x=π/6,-5/6π,-π/3,5/6π

熱心網友

1。f(x)=2sin(x+θ/2)cos(x+θ/2)+2√3cos(x+θ/2)^2-√3=sin(2x+θ)+√3(2cos(x+θ/2)^2-1)=sin(2x+θ)+√3cos(2x+θ)=2[(1/2)sin(2x+θ)+(√3/2)cos(2x+θ)]=2sin[2x+θ+(π/3)]2。由f(-x)=f(x),有2sin[-2x+θ+(π/3)]=2sin[2x+θ+(π/3)]，即：sin(-2x+θ)+√3cos(-2x+θ)=sin(2x+θ)+√3cos(2x+θ)sin(-2x+θ)-sin(2x+θ)=√3cos(2x+θ)-√3cos(-2x+θ)-sin2xcosθ=-√3sinθsin2xtanθ=√3/3∵0≤θ≤π，所以θ=π/63。此時f(x)=2sin(2x+π/6+π/3)=2sin(2x+π/2)=2cos2x=1,即cos2x=1/2x∈[-π,π], == 2x∈[-2π,2π],于是：2x=-2π+π/3,-π/3或2x=2π-π/3,π/3即：x={-5π/6,-π/6,π/6,5π/6}。

熱心網友

1)f(x)=2sin(x+θ/2)cos(x+θ/2)+2倍根號3倍的cos(x+θ/2)^2-根號3 =sin(2x+θ)+根號3*cos(2x+θ)=2sin(2x+θ+60').2)θ+60'=90',θ=30',f(x)=cos2x,為偶函數3)f(x)=cos2x=1,2x=2派,x=派.

熱心網友

cos(x+θ/2)^2-根號3看不懂啊也寫清楚點咯