求做一道三角函数题已知f(x)=sin(2x＋π/6)＋ cos(2x-π/3)＜1＞求f(x)的最大值及取得最大值时x

（1）f(x)=sin(2x+π/6)+cos(2x+π/6-π/2)
=sin(2x+π/6)+sin(2x+π/6)
=2sin(2x+π/6)
当2x+π/6=π/2时,即x=π/6时,f(x)最大为2
当2x+π/6=-π/2时,即x=-π/3时,f(x)最小为-2
（2）f(c)=1,
即f(c)=2sin(2C+π/6)=1
2C+π/6=5π/6
C=π/3
sinA=2sinB
sinA=2sin(π-A-C)
sinA=2sin(π-A-π/3)
sinA=2sin(A+π/3)=2(sinAcosπ/3+cosAsinπ/3)=sinA+2√3cosA
即2√3cosA=0,所以A=π/2
所以B=π-A-C=π-π/2-π/3=π/6
可以知道该三角形为指教三角形,A为直角,b、c为直角边
b=c*tanB=2√3*√3/3=2
S=b*c/2=2*2√3/2=2√3

最大值为2，此时x取30度。

f(x)=sin(2x＋π/6)＋ cos(2x-π/3)
=sin(2x＋π/6)+sin(2x+π/6)
=2sin(2x＋π/6)
1) ymax=2
2x＋π/6=2kπ+π/2
x=kπ+π/6 (k∈Z)
2) 2sin(2C＋π/6)=1
C=π/3
a/sinA=b/sinB
a=2...