设函数f(x)=cos(2x+π/3)+sin^2 X
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设函数f(x)=cos(2x+π/3)+sin^2 X
(1)求函数f(x)的最大值和最小正周期
(1)求函数f(x)的最大值和最小正周期
![设函数f(x)=cos(2x+π/3)+sin^2 X](/uploads/image/z/2899275-51-5.jpg?t=%E8%AE%BE%E5%87%BD%E6%95%B0f%28x%29%3Dcos%282x%2B%CF%80%2F3%29%2Bsin%5E2+X)
f(x)=cos2x*1/2-√3*sin2x+(1-cos2x)/2
=cos2x-√3sin2x+1/2
=2cos(2x+π/3)+1/2
所以最小正周期T=2π/2=π
当cos(2x+π/3)=1
取最大值
为5/2
=cos2x-√3sin2x+1/2
=2cos(2x+π/3)+1/2
所以最小正周期T=2π/2=π
当cos(2x+π/3)=1
取最大值
为5/2
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