Real Solutions of 4cos(e^x) = 2^x+2^-x: ln(2pi) < log2(2+sqrt3) < ln(3pi)

[juantheron]

If $\ln(2\pi)<\log_2(2+\sqrt{3})<\ln(3\pi)$, then find number of roots of the equation $$4\cos(e^x)=2^x+2^{-x}$$

 

[Albert1]

If $\ln(2\pi)<\log_2(2+\sqrt{3})<\ln(3\pi)$, then find number of roots of the equation $$4\cos(e^x)=2^x+2^{-x}$$

Spoiler

let $A=ln(2\pi)\approx 1.84,B=log_2(2+\sqrt 3)\approx 1.9,C=ln(3\pi)\approx 2.24,\,\, given: A<B<C$
let $f(x)=-2\leq 2\cos(e^x)\leq 2,g(x)=\dfrac {2^x+2^{-x}}{2}\geq 1$
for $g(x)$ is symmetric we only have to discuss $1\leq g(x)\leq2,\,\, or (\,\, -2\leq x\leq 2)$
some values for $-2\leq x\leq 2$ as bellow:

g(0)=1	f(0)≒1.08
g(0.5)≒1.06	f(0.5)≒-0.16
g(1)≒1.25	f(1)=-1.82
g(1.7)≒1.78	f(1.7)≒1.38
g(1.84)≒1.93	f(1.84)≒2
g(1.9)≒2	f(1.9)≒1.84
g(2)≒2.13	f(2)≒0.9
g(-1)≒1.25	f(-1)≒1.87
g(-2)≒2.13	f(-2)=1.98

from the above table we see there are four roots to the given equation
$0<root1<0.5$
$1.7<root2<1.84$
$1.84<root3<1.9$
$-2<root4<-1$