function y=sol(a1,a2)
syms x;
fun=a1*tan(a1*x)-a2*tan(a2*x);
y=solve(fun,'x');
sol(2,3)就可以得出如下结果：
ans =

[ 0]
[ pi]
[ atan((10-2*i*15^(1/2))^(1/2),(6+2*i*15^(1/2))^(1/2))]
[ atan(-1/4*(10-2*i*15^(1/2))^(1/2),1/4*(6+2*i*15^(1/2))^(1/2))]
[ atan(1/4*(10-2*i*15^(1/2))^(1/2),-1/4*(6+2*i*15^(1/2))^(1/2))]
[ atan(-1/4*(10-2*i*15^(1/2))^(1/2),-1/4*(6+2*i*15^(1/2))^(1/2))]
[ atan((10+2*i*15^(1/2))^(1/2),(6-2*i*15^(1/2))^(1/2))]
[ atan(-1/4*(10+2*i*15^(1/2))^(1/2),1/4*(6-2*i*15^(1/2))^(1/2))]
[ atan(1/4*(10+2*i*15^(1/2))^(1/2),-1/4*(6-2*i*15^(1/2))^(1/2))]
[ atan(-1/4*(10+2*i*15^(1/2))^(1/2),-1/4*(6-2*i*15^(1/2))^(1/2))] 

我想把上述方程解 正实数取出，然后具有正实部的部分取出来 ，其他的不要，用什么函数可以办到？ 请高手指教！