已知f (x)=2sin(x+θ2)cos(x+θ2)+23cos2(x+θ2)-3.
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已知f (x)=2sin(x+
θ |
2 |
(1)f(x)=sin(2x+θ)+2
3×
1+cos(2x+θ)
2-
3
=sin(2x+θ)+
3cos(2x+θ)
=2sin(2x+θ+
π
3);
(2)要使f (x)为偶函数,则必有f(-x)=f(x),
∴2sin(-2x+θ+
π
3)=2sin(2x+θ+
π
3),即-sin[2x-(θ+
π
3)]=sin(2x+θ+
π
3),
整理得:-sin2xcos(θ+
π
3)+cos2xsin(θ+
π
3)=sin2xcos(θ+
π
3)+cos2xsin(θ+
π
3)
即2sin2xcos(θ+
π
3)=0对x∈R恒成立,
∴cos(θ+
π
3)=0,又0≤θ≤π,
则θ=
π
6;
(3)当θ=
π
6时,f(x)=2sin(2x+
π
2)=2cos2x=1,
∴cos2x=
1
2,
∵x∈[-π,π],
∴x=±
π
6,
则x的集合为{x|x=±
π
6}.
3×
1+cos(2x+θ)
2-
3
=sin(2x+θ)+
3cos(2x+θ)
=2sin(2x+θ+
π
3);
(2)要使f (x)为偶函数,则必有f(-x)=f(x),
∴2sin(-2x+θ+
π
3)=2sin(2x+θ+
π
3),即-sin[2x-(θ+
π
3)]=sin(2x+θ+
π
3),
整理得:-sin2xcos(θ+
π
3)+cos2xsin(θ+
π
3)=sin2xcos(θ+
π
3)+cos2xsin(θ+
π
3)
即2sin2xcos(θ+
π
3)=0对x∈R恒成立,
∴cos(θ+
π
3)=0,又0≤θ≤π,
则θ=
π
6;
(3)当θ=
π
6时,f(x)=2sin(2x+
π
2)=2cos2x=1,
∴cos2x=
1
2,
∵x∈[-π,π],
∴x=±
π
6,
则x的集合为{x|x=±
π
6}.
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