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 Author Comment/Response Bill Simpson 05/28/12 6:42pm For sufficiently small and simple a and b. In[1]:= Simplify[a=5;b=3;Reduce[a*Sin[a x]-b*Sin[b x]==0,x]] Out[1]= C[1] ∈ Integers && (x == 2*Pi*C[1] || Pi + 2*Pi*C[1] == x || x + 2*ArcTan[Sqrt[Root[1 - 18*#1 + 42*#1^2 - 18*#1^3 + #1^4 & , 1, 0]]] == 2*Pi*C[1] || x == 2*(ArcTan[Sqrt[Root[1 - 18*#1 + 42*#1^2 - 18*#1^3 + #1^4 & , 1, 0]]] + Pi*C[1]) || x + 2*ArcTan[Sqrt[Root[1 - 18*#1 + 42*#1^2 - 18*#1^3 + #1^4 & , 2, 0]]] == 2*Pi*C[1] || x == 2*(ArcTan[Sqrt[Root[1 - 18*#1 + 42*#1^2 - 18*#1^3 + #1^4 & , 2, 0]]] + Pi*C[1]) || x + 2*ArcTan[Sqrt[Root[1 - 18*#1 + 42*#1^2 - 18*#1^3 + #1^4 & , 3, 0]]] == 2*Pi*C[1] || x == 2*(ArcTan[Sqrt[Root[1 - 18*#1 + 42*#1^2 - 18*#1^3 + #1^4 & , 3, 0]]] + Pi*C[1]) || x + 2*ArcTan[Sqrt[Root[1 - 18*#1 + 42*#1^2 - 18*#1^3 + #1^4 & , 4, 0]]] == 2*Pi*C[1] || x == 2*(ArcTan[Sqrt[Root[1 - 18*#1 + 42*#1^2 - 18*#1^3 + #1^4 & , 4, 0]]] + Pi*C[1])) In[2]:= ToRadicals[Root[1-18 #1+42 #1^2-18 #1^3+ #1^4&,1]] Out[2]= 9/2+Sqrt[41]/2-Sqrt[(59+9*Sqrt[41])/2] URL: ,

 Subject (listing for 'Can Mathematica solve this kind of equation?') Author Date Posted Can Mathematica solve this kind of equation? Fishbb 05/27/12 11:05pm Re: Can Mathematica solve this kind of equation? Bill Simpson 05/28/12 6:42pm Re: Re: Can Mathematica solve this kind of equa... Fishbb 05/29/12 10:30pm
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