作业帮已知函数f(x)=sin(2x+α)+根号3cos(2x+α)(0
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已知函数f(x)=sin(2x+α)+根号3cos(2x+α)(0
已知函数f(x)=sin(2x+α)+根号3cos(2x+α)(0
已知函数f(x)=sin(2x+α)+根号3cos(2x+α)(0
f(x)=sin(2x+α)+根号3cos(2x+α)
=2sin(2x+α+π/3)
∵f(x)图像过（π/12,1)
∴f(π/12)=2sin(π/6+π/3+α)=2sin(π/2+α)=2cosα=1
∴cosα=1/2
∵0
(1)f(x)=2(1/2sin(2x+a)+√3/2cos(2x+a))
=2sin(2x+a+π/3)
将点代入原函数，解得a=π/3
(2)由（1）得f(x)=2sin(2x+2π/3)
当-π/2+2kπ<(2x+2π/3)<π/2+2kπ为增区间。解得-7π/12+kπ
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(1)f(x)=2(1/2sin(2x+a)+√3/2cos(2x+a))
=2sin(2x+a+π/3)
将点代入原函数，解得a=π/3
(2)由（1）得f(x)=2sin(2x+2π/3)
当-π/2+2kπ<(2x+2π/3)<π/2+2kπ为增区间。解得-7π/12+kπ
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（1）π/3
（2）［-7π/12+kπ，-π/12+kπ］和［-π/12+kπ,5π/12+kπ］
第一题（1）先将原式转化f(x)=2sin(2x+α+π/3)（2）在将点（π/12,1)带入上式中得：2sin(π/2+α)=1（3）由于0<α<π/2，所以可以得出α=π/3
第二题（1）由第一题得出f(x)=2sin(2x+2π/3)（2）单调递增区间：首先-π/2<=2x+2π/3<=π/2再化简-7π/12<=x<=-π/12最后得出单调递增区间为：［-7π/12,-π/12］（...
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第一题（1）先将原式转化f(x)=2sin(2x+α+π/3)（2）在将点（π/12,1)带入上式中得：2sin(π/2+α)=1（3）由于0<α<π/2，所以可以得出α=π/3
第二题（1）由第一题得出f(x)=2sin(2x+2π/3)（2）单调递增区间：首先-π/2<=2x+2π/3<=π/2再化简-7π/12<=x<=-π/12最后得出单调递增区间为：［-7π/12,-π/12］（3）单调递减区间：首先π/2<=2x+2π/3<=3π/2再化简-π/12<=x<=5π/12最后得出单调递减区间为：［-π/12,5π/12］**最后最好总结一下
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