#e^(xcoshx)=3#

#xcoshx=ln3#

#x(e^x+e^-x)=ln9#

Let: #X=e^x#

#lnX(X+1/X)=ln9#

#(XlnX+1)/X=ln9#

#XlnX+Xln(e^(1/X))-Xln9=0#

#Xln((Xe^(1/X))/9)=0#

So:

#X=0#

or: #ln((Xe^(1/X))/9)=0#

#cancel(e^x=0)#, #e^x# is never equal to 0.

#e^x*e^(e^-x)/9=1#

#e^(x+e^x)=ln9#

#x+e^x=ln(ln(9))#

We cannot find the exact root, but because that #x+e^x-ln(ln(9))=0# is a continuous and monotonic function and between [-1;1] this function is negative and then positive, using the intermediate values theorem, we know that there's an unique #α in [-1;1]# where #e^(αcoshα)=0#

\0/ here's our answer!