Solving sin⁡(xx2+1)+sin⁡(1x2+x+2)=0Here is my way:First write the equation assin⁡(x2(x2+1)+12(x2+x+2))cos⁡(x2(x2+1)−12(x2+x+2))=0Then the real solution can be find byx2(x2+1)+12(x2+x+2)=0which gives x=-1.

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Ben Castillo · Accepted Answer

I do not know if this is more elegant.Let f(x)=sin⁡x,   g(x)=xx2+1 and h(x)=1x2+x+2Then your equation becomesf(g(x))+f(h(x))=0Since f(x)=sin⁡x is an odd function,−f(h(x))=f(−h(x))Then, your equation is thatf(g(x)t)=f(−h(x))It is known that in the range (−π2,π2) the function f(x) is increasing, and since −12≤g(x)≤12 and 0&amp;lt;h(x)≤47 your equation is equivalent tog(x)=−h(x)↔xx2+1=−1x2+x+2Therefore, the real solution comes fromx3+2x2+2x+1=0,which gives x=-1

Solving \(\displaystyle{\sin{{\left({\frac{{{x}}}{{{x}^{{2}}+{1}}}}\right)}}}+{\sin{{\left({\frac{{{1}}}{{{x}^{{2}}+{x}+{2}}}}\right)}}}={0}\) Here is my way: First write the

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