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I am unable to solve the following system of equations using Solve.

Solve[
  {1[E^(-i (k1 + k2)) (1 + E^(i (k1 + k2))) (E^(2 i k2) x + E^(2 i k1) y)] == 
     2 (-1 + Cos[k1] + Cos[k2])[E^(i k2) x + E^(i k1) y], 
   2 ((-(1/2))
     [E^(-i (k1 - 2 k2)) x + E^(2 i k1 - i k2) y + E^(3 i k2) (x + y)] + 
   (Cos[k1] + Cos[k2])[E^(2 i k2) x + E^(2 i k1) y]) == 
     1[(1 + E^(i (k1 + k2))) (E^(i k2) x + E^(I k1) y)]}, 
  {x, y}]`
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closed as off-topic by AccidentalFourierTransform, rhermans, Mariusz Iwaniuk, Michael E2, m_goldberg Jul 11 at 15:51

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question arises due to a simple mistake such as a trivial syntax error, incorrect capitalization, spelling mistake, or other typographical error and is unlikely to help any future visitors, or else it is easily found in the documentation." – AccidentalFourierTransform, rhermans, Mariusz Iwaniuk, Michael E2, m_goldberg
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 2
    $\begingroup$ If MMA cannot solve it, I'm not sure what you expect us to do. Solvable equations are the exception, not the rule. $\endgroup$ – AccidentalFourierTransform Jul 11 at 13:52
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    $\begingroup$ You have two problems: need upper case I for imaginary i and you can't use brackets to enclose expressions. Use parentheses. Make those changes and try it. $\endgroup$ – Dominic Jul 11 at 14:21
  • $\begingroup$ Some clarification: Use parenthesis to enclose expressions but brackets for function arguments like (a+b Cos[a]) and double brackets for array indexing like myArray[[index]]. $\endgroup$ – Dominic Jul 11 at 14:29
  • $\begingroup$ Hi user66523, welcome to Mma.SE. Start by taking the tour now and learning about asking and what's on-topic. Your question was put on-hold as it seems to be off-topic, i.e it arises from a simple mistake (syntax error) and will not help future visitors. Don't be discouraged by that cleaning-up policy. Your future good questions are welcome. Learn about common pitfalls here. Why not choosing a meaningful username? $\endgroup$ – rhermans Jul 11 at 16:09
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Replace i with I and some [] with () and simplify to see what we have.

Simplify[{1(E^(-I (k1 + k2)) (1 + E^(I (k1 + k2))) (E^(2 I k2) x + E^(2 I k1) y)) ==
  2 (-1 + Cos[k1] + Cos[k2])(E^(I k2) x + E^(I k1) y),
  2 ((-(1/2))(E^(-I (k1 - 2 k2)) x + E^(2 I k1 - I k2) y + E^(3 I k2) (x + y)) +
    (Cos[k1] + Cos[k2])(E^(2 I k2) x + E^(2 I k1) y)) == ((1 + E^(I (k1 + k2)))*
    (E^(I k2) x + E^(I k1) y))}]

(*{(1 - 2*E^(I*k2) + E^(I*(k1 + k2)))*x + (1 - 2*E^(I*k1) + E^(I*(k1 + k2)))*y == 0,
   E^((3*I)*k1)*y == E^((3*I)*k2)*y}*)

Now try to solve

Solve[%,{x,y}]

(*{{x->0,y->0}}*)

Using Reduce instead of Solve will show a much longer result with various possible conditions on some expressions which may or may not hold in your situation. If some of those conditions hold then there are other potential solutions.

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