若函数f（x）=asin（x-π/3）+b满足f（π/3）+f（π/2）=7且f（π）-f（0）=2√3求
问题描述：
若函数f（x）=asin（x-π/3）+b满足f（π/3）+f（π/2）=7且f（π）-f（0）=2√3求
（1）f（x）的解析式（2）f（x）的单调区间（3）f（x）的最小值（4）使f（x）=1/2的x的集合
答
1.f(π/3) + f(π/2）=[asin(π/3 - π/3) + b] + [asin(π/2 - π/3) + b] =(asin0 + b) + [asin(π/6) + b]=b + (1/2)a + b =7f(π) - f(0) =[asin(π - π/3) + b] - [asin(0 - π/3) + b] =[asin(2π/3) + b] - ...