Solve for x to three significant digits?

Question

#e^x + e^-x = 2#

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Answer 1 · 2016-12-13T15:14:45

Real solution: #x = 0#

Complex solutions: #x = 2kpi i# for any integer #k#

Let #t = e^x#

Then our equation becomes:

#t + 1/t = 2#

Mutliply through by #t# and subtract #2t# from both sides to find:

#0 = t^2-2t+1 = (t-1)^2#

So the only possible value of #t# is #t = 1#

Now solve #e^x = 1#

The unique Real solution is #x = 0# since #a^0 = 1# for all #a != 0#

There are also Complex solutions resulting from Euler's identity:

#e^(ipi) = -1#

Hence:

#e^((2kpi)i) = ((e^(ipi))^2)^k = ((-1)^2)^k = 1^k = 1#

for any integer #k#