Right side of the equation is a geometric series with "a = sin(x)" and "r = \\frac{sin^2(x)}{sin(x)} = sin(x)".
The sum of a geometric series:
Hence the equation is:
"sin(x^2-1)(1-sin(x)) = sin(x)(1-sin(x^2-1))"
"sin(x^2-1) = sin(x)"
Since the period the function "y = sin(x)" is "2\\pi", the equation is:
"x^2-x-(1+2\\pi n) = 0"
"x_1 = \\frac{1-\\sqrt{5+8\\pi n }}{2}, x_2 = \\frac{1+\\sqrt{5+8\\pi n }}{2}, n = 0, 1, 2, ..."
In particular, for n = 0 we have:
For n = 1:
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