Show that between any two roots of $e^x\cos(x)=1$ there exists atleast
one root of $e^x\sin(x)-1$
My Attempt:
Let $f(x)=e^x\cos(x)-1$ and $g(x)=e^{x}\sin(x)-1$
Let $a$ and $b$ be two consecutive roots of $f(x)$.
By Rolle's theorem, there exists a $c \in (a,b)$ such that $f'(c)=0$
$\implies e^c(\cos(c)-\sin(c))=0 \implies e^c\cos(c)-1=e^c\sin(c)-1$.
$c$ is not a root of $e^x\cos(x)-1$. So, I don't see how there can be a root of $e^{x}\sin(x)-1$ in $[a,b]$.
Am I going wrong somewhere?
Best Answer
Let $f(x)=\cos(x)-e^{-x}$.
If $a$ and $b$ are two consective roots of $f$ then there exists a $c\in(a,b)$ such that $-\sin(c)+e^{-c}=0$ from which the quoted claim follows.