$Find the real solution of the equation $x^{2}-10=x \sin{x}$.$

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$Find the real solution of the equation $x^{2}-10=x \sin{x}$.$

Does $x^{2}-10=x \sin{x}$ have a real solution? Solve the task without drawing a graph.
Explain the solution step by step, explain how to obtain the answer from the given information methodically, clearly, and coherently.

Calculus Trigonometry Precalculus

Egbonagba

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Define
\[f(x)=x^2-10-x\sin x.\]
Then
\[f(0)=0^2-10-0=-10<0\]
and
\[f(2\pi)=(2\pi)^2-10-2\pi \sin (2\pi)=4\pi^2-10>0.\]
Hence $f(0)<0$ and $f(2\pi)>0$. By the intermidiate value theorem, there exists $c$ in $(0,2\pi)$ such that
$f(c)=0$. Hence
\[0=f(c)=c^2-10-c\sin c \Rightarrow c^2-10=c\sin c.\]
Thus the given equation has a real solution.

Erdos

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$Find the real solution of the equation $x^{2}-10=x \sin{x}$.$

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