(2011•淮南一模)已知函数f(x)=cos2x−sin2x+23sinxcosx+1.
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(2011•淮南一模)已知函数f(x)=cos
(Ⅰ)f(x)=cos2x−sin2x+2
3sinxcosx+1=
3sin2x+cos2x+1
=2sin(2x+
π
6)+1. 由 2kπ+
π
2≤2x+
π
6≤2kπ+
3π
2,∴kπ+
π
6≤x≤kπ+
2π
3,
∴当k=-1时,∴−
5π
6≤x≤−
π
3; 当k=0时,∴
π
6≤x≤
2π
3,
又∵x∈[−
π
2,
π
3],∴ −
π
2≤x≤−
π
3,或
π
6≤x≤
π
3,
所以,函数f(x)单调减区间为:[−
π
2, −
π
3]和[
π
6,
π
3].
(Ⅱ)g(x)=2cos2x=2sin(2x+
π
2)=2sin2(x+
π
4),
把 f(x)=2sin(2x+
π
6)+1=2sin2(x+
π
12 ) 项左平移
π
6个单位,再向下平移1个单位,即得g(x)的解析式,
故
3sinxcosx+1=
3sin2x+cos2x+1
=2sin(2x+
π
6)+1. 由 2kπ+
π
2≤2x+
π
6≤2kπ+
3π
2,∴kπ+
π
6≤x≤kπ+
2π
3,
∴当k=-1时,∴−
5π
6≤x≤−
π
3; 当k=0时,∴
π
6≤x≤
2π
3,
又∵x∈[−
π
2,
π
3],∴ −
π
2≤x≤−
π
3,或
π
6≤x≤
π
3,
所以,函数f(x)单调减区间为:[−
π
2, −
π
3]和[
π
6,
π
3].
(Ⅱ)g(x)=2cos2x=2sin(2x+
π
2)=2sin2(x+
π
4),
把 f(x)=2sin(2x+
π
6)+1=2sin2(x+
π
12 ) 项左平移
π
6个单位,再向下平移1个单位,即得g(x)的解析式,
故
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