Question

已知函数f(x)=cos(x+x/6)-sin(x-2π/3)+sinx+a的最大值为1。求常数a的值? 求使f(x)≥0成立的x的取值范围

Answer 1

f(x) = cos(x+π/6)-sin(x-2π/3)+sinx+a
= cos(x+π/6)-{-sin(π+x-2π/3)} + sinx + a
= cos(x+π/6)+sin(x+π/3)} + sinx + a
= cos(x+π/6)+cos{π/2-(x+π/3)} + sinx + a
= cos(x+π/6)+cos{π/6-x} + sinx + a
= cos(x+π/6)+cos{x-π/6} + sinx + a
= cosxcosπ/6-sinxsinπ/6 + cosxcosπ/6+sinxsinπ/6 + sinx + a
= 2cosxcosπ/6 + sinx + a
= 2(cosxcosπ/6 + sinxsinπ/6) + a
= 2cos(x-π/6)+a
-2+a≤2cos(x-π/6)+a≤2+a
2+a=1
a=-1

Answer 2

f(x)=cos(x+x/6)-sin(x-2π/3)+sinx+a
=√3/2cosx-sinx/2+sinx/2+√3/2cosx+sinx+a
=√3cosx+sinx+a
=2sin(x+∏/3)+a
最大值2+a=1
a=-1
f(x)=2sin(x+∏/3)-1≥0
sin(x+∏/3)≥1/2
∏/6+2k∏≤x+∏/3≤5∏/6+2k∏
-∏/6+2k∏≤x≤∏/2+2k∏

Answer 3

cos(x+x/6）你确定不是cos(x+π/6）？

追问

啊 是cos(x+π/6）

追答

先将原式展开。
f(x)=跟号3/2cosx-1/2sinx+1/2sinx+根号3/2cosx+a
=2sin(x+π/3）+a
所以a=-1
f(x)=2sin(x+π/3)-1
依题意sin(x+π/3）≥1/2
以图像可得X∈【-π/6+2kπ，π/2+2kπ】