Finding The Number Of Solutions To $e^x -2 = \cos x$

Question

Hello everyone how many solution the equation $e^x -2 = \cos x$ have?

I tried to convert it to a function $y = e^x - \cos x -2$ and Find extreme points by:

$y' = e^x + \sin x$ but I don't know how to solve this equation.

Answer 1

Solutions can only be found where

$$-1\le e^x-2\le1,$$ i.e. $x\in[0,\log 3].$ In this range, $e^x-2$ is increasing and $\cos x$ decreasing, and there can only be one solution, which does exist, as

$$e^0-2<\cos0\land e^1-2>\cos1^*.$$

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$^*$By Taylor,

$$e-2>\frac12+\frac1{3!}=0.6666666\cdots$$ and $$\cos1<1-\frac12+\frac1{4!}=0.5416666\cdots$$

Answer 2

Just a visive approach with Desmos plotting two elementary functions $f(x)=\cos x$ and $g(x)=e^x-2$ (it is the $e^x$ shifted down of $-2$) and you can seen one only solution.