How many positive and negative roots of the equation $e^x-5\sin x$ exist?

Question

I know how to find positive and negative roots for polynomials using Descartes rule but I've no idea how to find the same for transcendental equations? Appreciate any help.

Answer 1

Since $0<e^x\ll 5$ when $x$ is large and negative, the function behaves essentially like $-5\sin x$ for large negative $x$, and so has $\infty$ negative roots.

Since $e^x\gg 5$ for large positive $x$, there are at most finitely many positive roots, none beyond $x=\ln 5$. You can plot it and see that it has two positive roots.

Answer 2

HINT: you must analyze the function using derivatives. You knows that the image of the function $e^x$ are the positive reals, and the image of the function $5\sin x$ is $[-5,5]$.

Thus the unique possible roots happen when $e^x\in (0,5)$ and $5\sin x\in(0,5)$. This imply that you must study the function in the domain $(-\infty,\ln 5)$. Because the sine function is periodic it cut $e^x$ infinite countable times when $x\le0$.

Thus there are infinite countable negative roots. It remains to see how many roots there are in the region $(0,\ln 5)$. Because $\ln 5\approx\pi/2$ then the sine function is almost injective in this region.