已知函数f(x)=5sin (x+π/3)-a[(cos x+π/3)^2]的图象经过点(-π/3,-2)

(1)点(-π/3,-2)代入函数式得
f(-π/3)=5sin 0-acos²0=-a=-2
所以a=2
(2)f(x)=5sin (x+π/3)-2cos²( x+π/3)
=5sin (x+π/3)-2+2sin ²(x+π/3)
=2[sin(x+π/3)+5/4]²-41/8
故当sin(x+π/3)=1时f(x)最大值为5
此时x+π/3=π/2+2kπ 即x的取值集合为{x|x=π/6+2kπ,k属于Z}
(3)定义域是[-π/2,π/2]时,x+π/3的范围是[-π/6,5π/6]
sin(x+π/3)的范围是[-1/2,1]
值域为[-4,5]