#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!
