呵呵~要会就两个都会~干吗会一个啊~(玩笑……)
f(x)=5sin(x+π/3)+2cos平方(x+π/3)
=5sin(x+π/3)+[2cos平方(x+π/3)+2sin平方(x+π/3)]-2sin平方(x+π/3)
=5sin(x+π/3)+2-2sin平方(x+π/3)
设sin(x+π/3)=t
f(x)=-2t平方+5t+2
=-2(t-5/4)平方+41/8 t∈[-1,1]
1、所以fmax=-2(1-5/4)平方+41/8=5
此时t=1 所以sin(x+π/3)=1
x取值集合：{x|x=2kπ+π/6(k∈Z)}
2、若定义域为[-π/2,π/2]
则t∈[-1/2,1/2]
fmax=-2(1/2-5/4)平方+41/8=4
fmin=-2(-1/2-5/4)平方+41/8=-1
函数的值域[4,-1]
祝你考试一路顺风~~~~~~