Question 366585: system of four linear algebraic equation (get values of A,B, C and D)
0= Acos(t)+Bsin(t)+Ctcos(t)+Dtsin(t)+3t-2
0=-Asin(t)+Bcos(t)+Ccos(t)-Ctsin(t)+Dsin(t)+Dtcos(t)+3
1=-Acos(t)-Bsin(t)-2Csin(t)-Ctcos(t)+2Dcos(t)-Dtsin(t)
1= Asin(t)-Bcos(t)-2Ccos(t)-Ccos(t)+Ctsin(t)-2Dsin(t)-Dsin(t)-Dtcos(t)
Answer by robertb(5830)   (Show Source):
You can put this solution on YOUR website! Let t = 0. Then the system becomes
0 = A-2,
0 = B+C+3,
1 = -A+2D,
1 = -B-2C-C, or 1 = -B - 3C.
The 1st equation gives A = 2.
Substitution into the 3rd equation gives 1 = -2 + 2D, or D = 3/2.
eliminating B from the 2nd and 4th equations gives C = 1.
Finally, substitution into the 2nd equation gives B=-4.
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