Sine Of Squared

This requires some subjective graphical analysis, (i.e., a bit of hand-waving) ....

Since $x^{2} \lt x$ for $0 \lt x \lt 1$ we have $\sin(x^{2}) \lt \sin(x)$ on this interval. The two curves intersect when $x = 1,$ after which, since $x^{2} \gt x$ for $x \gt 1,$ the curve $\sin(x^{2})$ "peaks" at $1$ when $x = \sqrt{\frac{\pi}{2}} \lt \frac{\pi}{2}.$ This curve then declines, intersecting $\sin(x)$ for some $x \lt \frac{\pi}{2}.$

Since $\sin(x) = \sin(\pi - x),$ this second point of intersection will thus occur when

$x^{2} = \pi - x \Longrightarrow x^{2} + x - \pi = 0 \Longrightarrow x = \boxed{\dfrac{-1 + \sqrt{1 + 4\pi}}{2}},$

where the positive root was taken as we are looking for $x \gt 1.$

$if\quad sin(A)=sin(B),\\ A=\quad \pi n+B\quad (or\quad \pi n-B)\quad \\ Here,\quad n\quad is\quad any\quad integer.\\ From\quad question,\quad sin(x)=sin({ x }^{ 2 })\quad So,\\ either{ \quad \quad x\quad =\quad x }^{ 2 }\quad \quad --1\quad \\ or\quad { x }^{ 2 }=\pi n-{ x }\quad --2\\ or\quad { x }^{ 2 }=\pi n+{ x }\quad --3\\ \\ Since\quad we\quad have\quad to\quad find\quad lowest\quad positive\quad solution,\quad we\quad take\quad n=1\\ in\quad eqn2\quad and\quad eqn3\quad (For\quad justification,\quad look\quad at\quad thier\quad discriminant).\\ \\ eqn1\quad gives\quad positive\quad solution\quad x=1\quad which\quad is\quad not\quad accepted\\ eqn2\quad gives\quad positive\quad solution\quad \frac { -1+\sqrt { 1+4\pi } }{ 2 } \quad \\ eqn3\quad gives\quad positive\quad solution\quad \frac { 1+\sqrt { 1+4\pi } }{ 2 } \\ \\ When\quad we\quad compare\quad solutions\quad from\quad eqn2\quad and\quad eqn3\\ we\quad see\quad \quad \quad \frac { -1+\sqrt { 1+4\pi } }{ 2 } \quad is\quad the\quad least\quad positive\quad value\quad of\quad x\\ greater\quad than\quad 1.\quad Hence\quad the\quad answer.$