作业帮已知函数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](/uploads/image/z/10195705-1-5.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3Dsin%282x%2B%CF%80%2F6%29%2B2%5Bsin%28x%2B%CF%80%2F6%29%5D%5E2-2%28cosx%29%5E2%2Ba-1%281%29%E6%9C%80%E5%B0%8F%E5%91%A8%E6%9C%9F%EF%BC%882%EF%BC%89%E5%8D%95%E8%B0%83%E9%80%92%E5%A2%9E%E5%8C%BA%E9%97%B4%EF%BC%883%EF%BC%89x%E5%B1%9E%E4%BA%8E%E3%80%900%2C%CF%80%2F2%E3%80%91%2Cf%28x%29min%3D1%2C%E6%B1%82a)
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