高1数学测试1已知f(x)=2s
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(...全部
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=-√3sinθsin2x
tanθ=√3/3
∵0≤θ≤π，所以θ=π/6
3。
此时f(x)=2sin(2x+π/6+π/3)=2sin(2x+π/2)=2cos2x=1,即
cos2x=1/2
x∈[-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}。
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