作业帮f(x)=2cos^2x+根号3sin2x+a(1)若X属于[0,π/2]时,f(x)最大值为4...f(x)=2cos^2x+√3sin2x+a(1)若X属于[0,π/2]时,f(x)最大值为4（1）若X属于R,求f(x)最小正周期（2）若X属于[0,π/2]时,fx的最大值是4,求a1,由辅助角公
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![f(x)=2cos^2x+根号3sin2x+a(1)若X属于[0,π/2]时,f(x)最大值为4...f(x)=2cos^2x+√3sin2x+a(1)若X属于[0,π/2]时,f(x)最大值为4（1）若X属于R,求f(x)最小正周期（2）若X属于[0,π/2]时,fx的最大值是4,求a1,由辅助角公](/uploads/image/z/8633303-71-3.jpg?t=f%28x%29%3D2cos%5E2x%2B%E6%A0%B9%E5%8F%B73sin2x%2Ba%281%29%E8%8B%A5X%E5%B1%9E%E4%BA%8E%5B0%2C%CF%80%2F2%5D%E6%97%B6%2Cf%28x%29%E6%9C%80%E5%A4%A7%E5%80%BC%E4%B8%BA4...f%28x%29%3D2cos%5E2x%2B%E2%88%9A3sin2x%2Ba%281%29%E8%8B%A5X%E5%B1%9E%E4%BA%8E%5B0%2C%CF%80%2F2%5D%E6%97%B6%2Cf%28x%29%E6%9C%80%E5%A4%A7%E5%80%BC%E4%B8%BA4%EF%BC%881%EF%BC%89%E8%8B%A5X%E5%B1%9E%E4%BA%8ER%2C%E6%B1%82f%28x%29%E6%9C%80%E5%B0%8F%E6%AD%A3%E5%91%A8%E6%9C%9F%EF%BC%882%EF%BC%89%E8%8B%A5X%E5%B1%9E%E4%BA%8E%5B0%2C%CF%80%2F2%5D%E6%97%B6%2Cfx%E7%9A%84%E6%9C%80%E5%A4%A7%E5%80%BC%E6%98%AF4%2C%E6%B1%82a1%2C%E7%94%B1%E8%BE%85%E5%8A%A9%E8%A7%92%E5%85%AC)
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f(x)=2cos^2x+根号3sin2x+a(1)若X属于[0,π/2]时,f(x)最大值为4...f(x)=2cos^2x+√3sin2x+a(1)若X属于[0,π/2]时,f(x)最大值为4（1）若X属于R,求f(x)最小正周期（2）若X属于[0,π/2]时,fx的最大值是4,求a1,由辅助角公
f(x)=2cos^2x+根号3sin2x+a(1)若X属于[0,π/2]时,f(x)最大值为4...
f(x)=2cos^2x+√3sin2x+a(1)若X属于[0,π/2]时,f(x)最大值为4
（1）若X属于R,求f(x)最小正周期
（2）若X属于[0,π/2]时,fx的最大值是4,求a
1,由辅助角公式可得：2sin(2x+π/6）+a+1
T=2π/ω=π
2,令2x+π/6=π/2,求出x等于π/6,即sin(2*π/6+π/6）=1
所以 2*1+a+1=4 所以a=1
这样做对吗?
f(x)=2cos^2x+根号3sin2x+a(1)若X属于[0,π/2]时,f(x)最大值为4...f(x)=2cos^2x+√3sin2x+a(1)若X属于[0,π/2]时,f(x)最大值为4（1）若X属于R,求f(x)最小正周期（2）若X属于[0,π/2]时,fx的最大值是4,求a1,由辅助角公
你是对的.
0
你是对的。