How do you solve #tanx=1# and find all exact general solutions?

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Answer 1 · 2018-03-02T19:38:52

#x in {((2n+1)pi)/2, n in ZZ}#

If #tan x = 1#, then #sin x = cos x#.

We know this is true for #x = pi/4# as a base case.

Now, the tangent function is periodic with it's period
#p = npi#, where #n# is an integer.

So #tan x = tan (x+npi)#. Go back to the original equality :

#tan x = tan (x+npi) = 1#

The first positive value #x_0# for which #tan x = 1# is, as stated before, #pi/4#.

#tan x_0 = tan(x_0 + npi) =1#

#tan(pi/4 + npi) = tan (((4n+1)pi)/4) = 1#

Which means #tan x = 1# if and only if #x# is a number of the form
#((4n+1)pi)/4#:

#x in {((4n+1)pi)/4 , n in ZZ}#.