已知函数f(x)=sin(2x+π/6)+2[sin(x+π/6)]^2-2(cosx)^2+a-1(1)最小周期（2）单调递增区间（3）x属于【0,π/2】,f(x)min=1,求a
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已知函数f(x)=sin(2x+π/6)+2[sin(x+π/6)]^2-2(cosx)^2+a-1(1)最小周期（2）单调递增区间（3）x属于【0,π/2】,f(x)min=1,求a
已知函数f(x)=sin(2x+π/6)+2[sin(x+π/6)]^2-2(cosx)^2+a-1
(1)最小周期（2）单调递增区间（3）x属于【0,π/2】,f(x)min=1,求a
已知函数f(x)=sin(2x+π/6)+2[sin(x+π/6)]^2-2(cosx)^2+a-1(1)最小周期（2）单调递增区间（3）x属于【0,π/2】,f(x)min=1,求a
化简;F(X)=√3*Sin2x-cos2x+a-1=2Sin（2x-π／6）+a-1,（1）最小周期T=π,（2) 单调递增区间：（-π／6+2kπ,π／3+2kπ）（3）x∈[0,π／2],x=0时,F﹙x﹚有最小值,F(x)min=a-2,又a-2=1,∴a=3.全手打.给分啊O(∩_∩)O!
解 f(x)=sin(2x+π/6)+[1-cos(2x+π/3)]-(1+cos2x)+a-1
=sin(2x+π/6)-cos(2x+π/3)-cos2x+a-1
=√3sin2x-cos2x+a-1=2sin(2x-π/6)+a-1
(1)最小周期T=2π/2=π
(2）根据正弦函数的性质，2kπ-π/2<=2x-π/6<=2kπ+π/2时函数递增。
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解 f(x)=sin(2x+π/6)+[1-cos(2x+π/3)]-(1+cos2x)+a-1
=sin(2x+π/6)-cos(2x+π/3)-cos2x+a-1
=√3sin2x-cos2x+a-1=2sin(2x-π/6)+a-1
(1)最小周期T=2π/2=π
(2）根据正弦函数的性质，2kπ-π/2<=2x-π/6<=2kπ+π/2时函数递增。
即kπ-π/6<=x<=kπ+π/3，k为整数
所以f(x)的单调递增区间为[kπ-π/6,kπ+π/3]. k为整数
(3)x属于【0，π/2】，-π/6<=2x-π/6<=5π/6
f(x)min=f(-π/6)=2(-1/2)+a-1=a-2=1
a=3
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已知f(x)=-1/2+sin(π/6-2x)+cos(2x-π/3）+cos平方x。⑴求f(xsin2x+[2(cosx)^2-1]/2 =cos2x+(cos2x)/2 =(3/2)cos2x 所以其最