Let's suppose "\\sin(x) \\neq \\pm1, \\sin(x^2-1)\\neq1" (as if it is not the case either the right side or the left side diverges). In this case the expression on the right converges to :
"\\sum_{k\\geq 1} \\sin^k(x) = \\frac{\\sin(x)}{1-\\sin(x)}" by the geometric progression sum formula.
Therefore we have :
"\\frac{\\sin(x^2-1)}{1-\\sin(x^2-1)}=\\frac{\\sin(x)}{1-\\sin(x)}"
As we need to find at least one solution, we will not seek the general solution of this equation and we will study at least the case :
"x^2-1=x" (as the solutions of this equations are also solutions of our equation)
"x_{+,-}=\\frac{1\\pm\\sqrt{1+4}}{2} = \\frac{1}{2}\\pm \\frac{\\sqrt{5}}{2}"
We know that both "x_{+,-}\\neq \\pm \\frac{\\pi}{2}" , so this solution satisfies our condition "\\sin(x),\\sin(x^2-1) \\neq \\pm 1". And thus we have found at least one (even two) solutions.
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