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Given that $$f(x)= 1 - e^x\sin(x)$$ $$g(x)= 1 + e^x\cos(x)$$

Using Rolle's theorem, prove that there exists at least one root of $g$ between any two roots of $f$. Attempt so far:

$f'(x) = -e^x(\sin(x)+\cos(x))= f(x)-g(x)$

Since $f(x)$ and $g(x)$ are continuous and differentiable, Rolle's theorem is applicable. Let $a$ and $b$ be any $2$ roots of $f$. Then by Rolle's theorem, there exists a $c$ in $(a,b)$ such that $f'(c) =0$. Therefore,$$f'(c)=0=f(c)-g(c)$$ $$\implies f(c)=g(c)$$

That is where I got stuck and I don't know how to proceed. Am I on the right track? What else should I be doing? Am I at all on the right track?

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