Question

Show that between any two roots of the equation $e^x\cos x=1$ there exists atleast one root of $e^x\sin x-1=0$ by continuity and differentiability.

Answer 1

Let $a$ and $b$ be the roots of the equation $e^x\cos x=1$ then

$e^a\cos a=1$ and $e^b\cos b=1$

Let $f$ be a function defined as $f\left(x\right)=e^{-x}-\cos x$

We observe that:

$\left(i\right)f\left(x\right)$ is continuous in $[a,b]$ and $\cos x$ both are continuous functions.

$\left(ii\right)f^{\prime }\left(x\right)=-e^{-x}+\sin x$, hence function is differentiable in $(a,b)$

$\left(iii\right)f\left(a\right)=e^{-a}-\cos a=e^{-a}(1-e^a\cos a)=0$ and

$f\left(b\right)=e^{-b}-\cos b=e^{-ab}(1-e^b\cos b)=0$

$\Rightarrow f\left(a\right)=f\left(b\right)=0$

Thus $f\left(x\right)$ satisfies all the conditions of Rolle's theorem in $[a,b]$

hence there exists at least one value of $x$ in $(a,b)$ such that $f^{\prime }\left(c\right)=0$

$f^{\prime }\left(x\right)=-e^{-x}+\sin x$

$f^{\prime }\left(c\right)=-e^{-c}+\sin c$

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

$\Rightarrow e^{-c}=\sin c$

$\Rightarrow e^c\sin x-1=0$

Thus, $c$ is a root of the equation $e^{-x}-\sin x=0$

Hence between any two roots of the equation $e^x\cos x=1$ there exists atleast one root of $e^x\sin x-1=0$