Inspired by Nihar Mahajan
Let be the sum of the integral values of that satisfies the ecuation and the number of solutions of the ecuation . Find
The answer is 0.
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1 solution
a and b are both zero. Why a is zero? The given interval contains 13 integral values, including zero. -6, 6, -5, 5, -4, 4..., -1,1, and 0. If value n satisfies the first equation, then value -n satisfies as well, because cosine is an even function and sin is an odd function. abs( sin(n)) = sin(n) or -sin(n). Therefore, cos(sin(n)) = cos(-sin(n)) = cos(sin(-n))= cos(abs(sin(-n))). So the total of all integral roots is 0 or a=0. b is zero, because the the second equation does not have roots. Let's work with the difference and convert it into the product: sin(cos(x)) - cos(sin(x)) = cos(Pi/2 - cos(x)) - cos(sin(x)) = -2 * sin((Pi/2 - cos(x) + sin(x))/2) * sin ((Pi/2 - cos(x) - sin(x))/2). The second equation has a solution only if this product turns to 0 at some x. The first factor is equal to 0 when (Pi/2 - cos(x) +sin(x))/2 = Pi * n, where n is an integer. The second factor is equal to 0 when (Pi/2 - cos(x) - sin(x))/2 = Pi * n, where n is an integer. But abs(sin(x) + cos(x)) or abs(sin(x) - cos(x)) does not exceed sq.root(2). And the right side even in case n=0 still will be Pi/2 > sq.root(2), which means that the second equation does not have a solution and b = 0. Finally, a + b = 0 is an answer.