(2012•甘肃一模)(文科)已知函数f(x)=3sin2x+23sinxcosx+cos2x,x∈R.
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(2012•甘肃一模)(文科)已知函数f(x)=3sin
(1)∵函数f(x)=3sin2x+2
3sinxcosx+cos2x=1+2sin2x+
3sin2x=1+1-cos2x+
3sin2x
=2+2(
3
2sin2x-
1
2cos2x)=2+2sin(2x-
π
6).
故当 sin(2x-
π
6)=1时,函数f(x)取得最大值为4.
令 2kπ-
π
2≤2x-
π
6≤2kπ+
π
2,k∈z,求得 kπ-
π
6≤x≤kπ+
π
3,
故函数的增区间为[kπ-
π
6≤xkπ+
π
3],k∈z.
(2)由f(x)≥3可得,sin(2x-
π
6)≥
1
2,
∴2kπ+
5π
6≥2x-
π
6≥2kπ+
3sinxcosx+cos2x=1+2sin2x+
3sin2x=1+1-cos2x+
3sin2x
=2+2(
3
2sin2x-
1
2cos2x)=2+2sin(2x-
π
6).
故当 sin(2x-
π
6)=1时,函数f(x)取得最大值为4.
令 2kπ-
π
2≤2x-
π
6≤2kπ+
π
2,k∈z,求得 kπ-
π
6≤x≤kπ+
π
3,
故函数的增区间为[kπ-
π
6≤xkπ+
π
3],k∈z.
(2)由f(x)≥3可得,sin(2x-
π
6)≥
1
2,
∴2kπ+
5π
6≥2x-
π
6≥2kπ+
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