How do we prove that $x^2=x\sin{x}+\cos{x}$ for exactly two values of $x$?
I thought of the sandwich theorem or such, getting only:
$-1\leq\sin{x}\leq1$
$-x\leq x\sin{x}\leq x$
$-x+\cos{x}\leq x\sin{x}+\cos{x}\leq x+\cos{x}$
Then replace the expression in the middle with $x^2$ because that is what says in the first equality:
$-x+\cos{x}\leq x+\cos{x}$
Then I got the idea to get the derivative for all three sides:
$-1-\sin{x}\leq2x+\sin{x}\leq1-\sin{x}$
Now we sum $\sin{x}$ in all three sides:
$-1\leq2x+\sin{x}\leq1$
That says that for f'(x), $x$ is zero.
However, I don't even know if I am doing well. What is next? Can you explain to me?