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

Let $f(x)=e^{-x}-\sin x$ and let $\alpha$ and $\beta$ be two roots of the equation $e^x \sin x-1=0$ such that $\alpha<\beta$. Then,

$e^\alpha \sin \alpha=1$ and $e^\beta \sin \beta=1$

$\Rightarrow e^{-\alpha}-\sin \alpha=0$ and $e^{-\beta}-\sin \beta=0$       ....(i)

Clearly, $f(x)$ is continuous on $[\alpha, \beta]$ and differentiable on $(\alpha, \beta)$.

Also, $f(\alpha)=f(\beta)=0$          [Using (i)]

Therefore, by Rolle's theorem there exists $c \in(\alpha, \beta)$ such that

$f^{\prime}(c)=0$

$\Rightarrow -e^{-c}-\cos c=0$

$\Rightarrow e^{-c}+\cos c=0$

$\Rightarrow e^c \cos c+1=0$

$\Rightarrow x=c$ is a root of $e^x \cos x+1=0$, where $c \in(\alpha, \beta)$