There are some steps:
[Step1] Prove that e^xcosx=1 has no negative root.
If x<0, the following must be satisfied:
0< e^x <1, -1<= cosx <=1
-> abs(e^xcosx)<1
So, the roots e^xcosx=1 must be 0 or greater.
[Step2] Find where the roots of e^xcosx=1 lie.
Let f(x)=e^xcosx-1.
f(0)=e^0*cos0-1=1*1-1=0 and x=0 is the smallest root.
Differentiate f(x):
f'(x)=e^x(cosx-sinx)
=-sqrt(2)e^xsin(x-1/4pi)
x=(n+1/4)pi satisfies f'(x)=0 if n>=0 is an integer.
Indeed, f((2n+1/4)pi) is a local maxima and f((2n+5/4)pi) is a local minima.
Here is a graph for y=e^xcosx-1.
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Let x=alpha is a positive root for f(x)=0.
alpha must satisfy (2n+1/4)pi< alpha< (2n+1/2)pi or (2n+3/2)pi< alpha < (2n+2)pi since:
f((2n+1/4)pi)=e^((2n+1/4)pi)/sqrt(2)-1> 0 (Note that e^(pi/4)=2.193... is larger than sqrt(2))
f((2n+1/2)pi)=-1< 0
f((2n+3/2)pi)=-1<0
f((2n+2)pi)=e^((2n+2)pi)-1> 0
There are no roots between 2npi< x<= (2n+1/4)pi because f(x) is monotonically increasing in this range(f'(x)> 0).
Similarly, there are no roots between (2n+1/2)pi<= x <=(2n+3/2)pi. f(x) is always negative here.
So,
color(red)("it is sufficient to prove that at least one root for " e^xsinx-1=0 " exists in the range " 2npi< x< (2n+1/4)pi or (2n+1/2)pi < x< (2n+3/2)pi.)
[Step3] Prove the statement above(in red characters)
Let g(x)=e^xsinx-1.
g(2npi)=-1< 0
g((2n+1/4)pi)=e^((2n+1/4)pi)/sqrt(2)-1> 0
g((2n+1/2)pi)=e^((2n+1/2)pi)-1> 0
g((2n+1)pi)=-1 <0
The proof is completed via the intermediate value theorem.
Visualization:
![enter image source here]()
Roots for e^xcosx-1=0 is in the red area and roots for e^xsinx-1=0 is in the green area.
