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我试图解决一个假设自变量为正值的三角函数。然而,Maxima 似乎没有考虑到求解例程的这种假设。
assume(t >0); solve(sin(t) = 0);
预期结果:
[t=%pi]
我得到什么:
[t=0]
一个可以使用to_poly_solve包;请参阅文档。
to_poly_solve
load(to_poly_solve); to_poly_solve(sin(t), t);
产生
%union([t = %pi %z0])
这是您期望的结果。