((e^x)+(e^-x))/2=1

This is an exponential equation and we'll solve it using
the substitution technique.

First, we'll note that we have
a negative power:

e^-x =
1/e^x

Now, we'll note e^x =
t

We'll re-write the equation in
t:

(t + 1/t) / 2 = 1

We'll
cross multiply:

t + 1/t =
2*1

t + 1/t = 2

We'll multiply
by t;

t^2 + 1 = 2t

We'll
subtract 2t both sides:

t^2 - 2t + 1 =
0

But the expression is resulted after expanding the
square

(t-1)^2 = 0

We'll put
t-1 = 0

t = 1

But e^x = t, so
e^x = 1 (1)

We could write 1 =
e^0.

We'll re-write (1):

e^x =
e^0

x = 0

The
solution of the equation is x = 0.