Assuming that the balance is a common two-pan one (rather than a digital one), here’s a possible solution with MAX. 3 attempts:

Step 1: Split the coins into two sets of three coins each.
We have TWO possibilities: One, BOTH counterfeit coins are in the SAME set (Case 1) OR both sets contain one counterfeit each (Case 2).
Step 2: Weigh the sets against each other. (W1).

Case 1: The set containing the counterfeits will be lighter. In this case, take this set for the next step (to identify the counterfeits). Let the coins be C11, C12 and C13.
Step 3(1): Keep one coin (C11) aside. Now, since two of the three coins are counterfeits, we have two possibilities here. P1: Both the remaining coins (C12 AND C13) are counterfeit; P2: C11 and one of the other two coins (C12 OR C13) are counterfeit.
Step 4(1) W2: Weigh C12 and C13. If they are equal, both are counterfeit. Otherwise, the lighter of the two and C11 the counterfeits. SOLVED.

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Case 2: Both the sets are of equal weight (because both contain one counterfeit each). Take any set. Let the coins be C21, C22, C23.
Step 3(2): Keep one coin (C21) aside. Again, we have two possibilities here: p1: the coin set aside (C21) is counterfeit; p2: One of the other two (C22 OR C23) is the counterfeit.

Step 4(2): W2: Weigh C22 and C23. If they are equal, both are good coins and C21 is the counterfeit. Otherwise, the lighter of the two (C22 / C23) is the counterfeit in the currently selected set. One of the counterfeits identified.

Step 5(2): W3: Repeat the Step 3(2) and 4(2) on the second set to determine the counterfeit in that set. BOTH Counterfeits identified; problem SOLVED.