Question

已知f(x)=2cos^2x+根号3sin2x+a(1)若X属于[0,π/2]时,f(x)最小值为-4,求a

详细过程。谢谢。

Answer 1

f(x)=2(cosx)^2+√3sin2x+a
=cos2x+1+√3sin2x+a
=2sin(2x+π/6)+a+1

x∈[0,π/2]
则π/6≤2x+π/6≤7π/6

所以f(x)的最小值是2*(-1/2)+a+1=a

所以a=-4

Answer 2

f(x)=2cos^2x+√3sin2x+a
=2*(1+cos2x)/2+√3sin2x+a
=1+cos2x+√3sin2x+a
=cos2x+√3sin2x+a+1
=2*(1/2*cos2x+√3/2*sin2x)+a+1
=2*(sinπ/6*cos2x+cosπ/6sin2x)+a+1
=2*sin(π/6+2x)+a+1
x∈[0,π/2]
2x∈[0,π]
π/6+2x∈[π/6,7π/6]
-1/2<=sin(π/6+2x)<=1
-1<=2sin(π/6+2x)<=2

当2sin(π/6+2x)=-1时，f(x)值最小
-4=-1+a+1
a=-4

Answer 3

不知道a(1)是什么
不过2cos^2x+根号3sin2x
=2cos²2x-1+根号3sin2x+1
=cos2x+根号3sin2x+1
=2(cos2xsinπ/6+sin2xsinπ/6)+1
=2cos(2x+π/6）+1
可以求出最小值，你在对照你的题看一下

Answer 4

f(x)=2cos^2x+根号3sin2x+a
f(x)=2cos^2x+√3sin2x+a
f(x)=1+cos2x+√3sin2x+a
f(x)=2sin(2x+π/6)+1+a
π/6<=2x+π/6<=7π/6
x=7π/6,最小
a=-4