Rolle's theorem on $e^x\sin x-1=0$

Question

Prove that between any two real roots of the equation $e^x\sin x-1=0$ the equation $e^x\cos x+1=0$ has at least one root.

My attempts:

By Rolle's theorem, the derivative of $e^x\sin x-1=0$ has at least one root between the roots $(\text{let}\ a,b)$ of $e^x\sin x-1=0$

$$f'(x)=e^x\sin x +e^x\cos x=\overbrace{\underbrace{(e^x\sin x-1)}_{2). \text{ no root in}\ (a,b)} +\underbrace{(e^x\cos x+1)}_{1,2\implies\text{ no root in}\ (a,b)}}^{\text{1). at least one root in}\ (a,b)}=0,\ x\in(a,b)$$

I found alternative proof of this here, but please don't mark this as duplicate as I want to know what's the wrong with what is tried.

Answer 1

$$f'(x)=e^x\sin x +e^x\cos x=\overbrace{\underbrace{(e^x\sin x-1)}_{2). \text{ no root in}\ (a,b)} +\underbrace{(e^x\cos x+1)}_{1,2\implies\text{ no root in}\ (a,b)}}^{\text{1). at least one root in}\ (a,b)}=0,\ x\in(a,b)$$

$$\text{1). at least one root in}\ (a,b)$$

$$\text{2). no root in}\ (a,b)$$

the whole expression has one root in $(a,b)$

the first expression has no root in $(a,b)$

then how will you get that at least one root that you are claiming for the whole expression.

So the 2nd expression gives us that root thus $$1,2\implies\text{ at least one root in}\ (a,b)$$

Answer 2

Refer to the graph:

$\hspace{2cm}$

You are mixing up the points $c$ with $d$.

By Rolle's theorem, the derivative of $e^x\sin x−1=0$ has at least one root between the roots (let $a,b$) of $e^x\sin x−1=0$.

True, let's call this point $d\in (a,b)$, then: $$f'(d)=e^d\sin d +e^d\cos d=(\underbrace{e^d\sin d-1}_{\ne 0})+(\underbrace{e^d\cos d+1}_{\ne 0})=0$$ However, the equation $e^x\cos x+1=0$ may (in fact, it does) have its root at the point $c\in (a,b)$, $c\ne d$. So, the above condition must look like: $$f'(d)=e^d\sin d +e^d\cos d=\\ (\underbrace{e^d\sin d+e^d\cos d-e^c\cos c-1}_{= 0})+(\underbrace{e^c\cos c+1}_{= 0})=0.$$ However, the equality of the first term to zero must be proved.

Note that the solution by zhw given in the link is efficient. Indeed, note how skillfully the exponential and sine functions are separated: $$f(x)=1-e^x\sin x = 0 \iff e^x\sin x=1 \iff \\ \sin x=e^{-x} \iff e^{-x}-\sin x=0.$$