1996 AIME Problems/Problem 10
Find the smallest positive integer solution to .
Since , multiplying both sides by yields .
Therefore, the smallest positive solution is .
which is the same as .
So , for some integer . Multiplying by gives . The smallest positive solution of this is
Solution 3 (Only sine and cosine sum formulas)
It seems reasonable to assume that for some angle . This means for some constant . We can set .Note that if we have equal to both the sine and cosine of an angle, we can use the sine sum formula (and the cosine sum formula on the denominator). So, since , if we have from the sine sum formula. For the denominator, from the cosine sum formula, we have This means so for some positive integer (since the period of tangent is ), or . Note that the inverse of modulo is itself as , so multiplying this congruence by on both sides gives For the smallest possible , we take
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