高一数学题，求解
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(1)振幅A=3，周期T=2π/2=π
（2）0≤x≤π/2 ==>0≤2x≤π ==>0-π/3≤2x-π/3≤π-π/3
==>-π/3≤2x-π/3≤2π/3
==>-1/2≤f(x)≤1 ==>f(x)min=-1/2, 2x-π/3=2π/3 x=π/2
当f(x)min=-1/2时，x=π/2
f(x)mxa=1, 2x-π/3=0 ,x=π/6
当f(x)mxa=1时，x=π/6
(3)f(a)+f(π/2)=3cos(2a-π/3)+3cos(2*π/2-π/3)=0
==>cos(2a-π/3)=-cos(2π/3)=1/2
2a=2kπ+π/3±π/3
所以：a=kπ 或者a=kπ+π/3
（2）0≤x≤π/2 ==>0≤2x≤π ==>0-π/3≤2x-π/3≤π-π/3
==>-π/3≤2x-π/3≤2π/3
==>-1/2≤f(x)≤1 ==>f(x)min=-1/2, 2x-π/3=2π/3 x=π/2
当f(x)min=-1/2时，x=π/2
f(x)mxa=1, 2x-π/3=0 ,x=π/6
当f(x)mxa=1时，x=π/6
(3)f(a)+f(π/2)=3cos(2a-π/3)+3cos(2*π/2-π/3)=0
==>cos(2a-π/3)=-cos(2π/3)=1/2
2a=2kπ+π/3±π/3
所以：a=kπ 或者a=kπ+π/3
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