Location of roots of certain transcendental equations. If $a$ and $b$ are two solutions of $\,e^x \cos x -1=0$, then how many solutions of the equation $ e^x \sin x-1=0$ lie between $a$ and $b$ ?  
 A: One solution of $f(x)=e^x\cos x-1=0$ is $x=0$ and there is clearly no solution with $x<0$. Between any two solutions, there must be a root of $f'(x) =(-\sin x+\cos x) e^x=\sqrt 2\sin(x-\frac\pi4)e^x$. Thus the first positive root is at $x\ge \frac\pi4$, so that $f(x)=0$ implies $|\cos x|\le e^{-\pi/4}<\frac12$ and hence there is $k\in\mathbb Z$ with $|x-((k+\frac12)\pi|<\frac\pi6$. Also, we have $f'(0)=1$ and for positive roots of $f$, $|\cos x|<\frac12$ implies $|\sin x|>\frac12$, hence $f'(x)\ne 0$, i.e. all solutions of $f(x)=0$ have multiplicity one.
Similarly, we find for $g(x)=e^x\sin x-1$ that all solutions of $g(x)=0$ are positive and between two such solutions must be a point with $g'(x)=0$, i.e. $(\sin x+\cos x)e^x=\sqrt 2\sin(x+\frac\pi 4)e^x=0$, hence the second positive solution of $g(x)=0$ is $\ge\frac{3\pi}4$ so that for all except the first roots of $g(x)$ we have $|\sin x|\le e^{-3\pi/4}<\frac12$, hence $|x-k\pi|<\frac16\pi$ for some $k\in\mathbb Z$. The first root on the other hand, is clearly between $x=0$ ($g(x)=-1$) and $x=\frac\pi4$ ($g(x)=e^{\pi/4}>1$).
In short: Apart from the first roots, the roots of $f(x)$ and $g(x)$ are relatively close to the well-behaved roots of $\cos$ and $\sin$ so that we conclude that the number of roots of $g$ between two numbers $0<a<b$ with $f(a)=f(b)=0$ is obtained by rounding $\frac{b-a}\pi$ to the nearest integer. If $a=0<b$, the number is obtained by rounding $\frac b\pi+\frac12$ to the nearest integer.
