Risolvi tanx = cotx per tutte le soluzioni [0, 2pi)?

Note that the initial presence of #tan(x) = sin(x)/cos(x)# e #cot(x) = cos(x)/sin(x)# implies we must have #sin(x)!=0# e #cos(x)!=0#. With that:

#tan(x) = cot(x)#

#=> sin(x)/cos(x) = cos(x)/sin(x)#

#=> sin(x)/cos(x)*sin(x)cos(x) = cos(x)/sin(x)*sin(x)cos(x)#

#=> sin^2(x) = cos^2(x)#

#=> sin(x) = +-cos(x)#

If we examine a unit circle, we find that this equality holds at #x=pi/4+npi/2, n in ZZ#. Thus, we must only find which values of #n# causare #x# to lay within the interval #[0, 2pi)# Testing, we find that

#pi/4+npi/2 in [0, 2pi)# for #n in {0, 1, 2, 3}#

Substituting those in, we get our answers:

#x in {pi/4, (3pi)/4, (5pi)/4, (7pi)/4}#