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Similar to yet distinct from the other solutions. Substitute two values for t (not a period apart), to create two independent equations. Then you will get a unique solution. Solve[-2 Exp[t] ((n + 3 m) Cos[2 t] + (-3 n + m) Sin[2 t]) == Exp[t] Cos[2 t] /. {{t -> 0}, {t -> 1}}, {n, m}] (* {{n -> -(1/20), m -> -(3/20)}} *) (Of course, the ...


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Here is another way to solve problems of this type: create a system of at least two independent equations for the two variables a and b. How to do this if you only have one equation to begin with? In your case one side of the equation (Exp[t] Cos[2 t]) is independent of the unknowns, so I just evaluate that side at some specific values of t and use those as ...


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You want to extract coefficients of terms in Cos[t] and other functions of t and find values of {A, B} that make these terms vanish. SolveAlways can do this sometimes (works reliably when input is polynomial, say). SolveAlways[ TrigExpand@{-2 Exp[t] ((A + 3 B) Cos[2 t] + (-3 A + B) Sin[2 t]) == Exp[t] Cos[2 t]}, {Sin[t], Cos[t]}] During ...



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