作业帮已知函数f(x)=2cosxsin(x+π/3)-根号3sin^2x+sinxcosx,当x∈[π/12,7π/12]时求f-1（1）的值
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![已知函数f(x)=2cosxsin(x+π/3)-根号3sin^2x+sinxcosx,当x∈[π/12,7π/12]时求f-1（1）的值](/uploads/image/z/2692367-71-7.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3D2cosxsin%28x%2B%CF%80%2F3%29-%E6%A0%B9%E5%8F%B73sin%5E2x%2Bsinxcosx%2C%E5%BD%93x%E2%88%88%5B%CF%80%2F12%2C7%CF%80%2F12%5D%E6%97%B6%E6%B1%82f-1%EF%BC%881%EF%BC%89%E7%9A%84%E5%80%BC)
已知函数f(x)=2cosxsin(x+π/3)-根号3sin^2x+sinxcosx,当x∈[π/12,7π/12]时求f-1（1）的值
已知函数f(x)=2cosxsin(x+π/3)-根号3sin^2x+sinxcosx,当x∈[π/12,7π/12]时求f-1（1）的值
已知函数f(x)=2cosxsin(x+π/3)-根号3sin^2x+sinxcosx,当x∈[π/12,7π/12]时求f-1（1）的值
已知函数f(x)=2cosxsin(x+π/3)-根号3sin^2x+sinxcosx,当x∈[π/12,7π/12]时求f-1（1）的值
解析：∵函数f(x)=2cosxsin(x+π/3)-√3sin^2x+sinxcosx
=2cosx(1/2sinx+√3/2cosx)-√3/2(1-cos2x)+1/2sin2x
=1/2sin2x+√3/2(1+cos2x)-√3/2(1-cos2x)+1/2sin2x
=sin2x+√3cos2x=2sin(2x+π/3)
∴f(x)=2sin(2x+π/3),其定义域为[π/12,7π/12]
∴值域为[-2,2]
其反函数为f^(-1)(x)=[ π-arcsin(x/2)- π/3]/2,其定义域为[-2,2],值域为[π/12,7π/12]
f^(-1)(1)=[ π-arcsin(1/2)- π/3]/2=π/4
f(x)=cosx(sinx+√3cosx)-√3sin2 x+sinxcosx
=2sinxcosx+√3(cos2 x-sin2 x)
=sin2x+√3cos2x
=2sin(2x+π/3)
=1
2x+π/3=π/6+2kπ或5π/6+2kπ
x∈[π/12，7π/12]
x=π/4
f（x）=2cosx（1/2·sinx+根号3/2·cosx）-根号3·（1-cos2x）/2+1/2·sin2x
=1/2·sin2x+根号3·（1+cos2x）/2-根号3·（1-cos2x）/2+1/2·sin2x
=sin2x+根号3·cos2x
=2（sin2xcosπ/3+cos2xsinπ/3）
=2si...
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f（x）=2cosx（1/2·sinx+根号3/2·cosx）-根号3·（1-cos2x）/2+1/2·sin2x
=1/2·sin2x+根号3·（1+cos2x）/2-根号3·（1-cos2x）/2+1/2·sin2x
=sin2x+根号3·cos2x
=2（sin2xcosπ/3+cos2xsinπ/3）
=2sin（2x+π/3）
求f-1（1）的值，即为求f（x）=1时，x的值
2sin（2x+π/3）=1时，x=-π/6+kπ或π/2+kπ
x∈[π/12，7π/12]时，x=π/2
即f-1（1）的值的值为π/2
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