Question 1109425
{{{cot(2x)=cos(2x)/sin(2x)}}} must be defined and must not be zero.
So, the restrictions will arise from
{{{cos(2x)<>0}}} and from {{{sin(2x)<>0}}} .
When {{{sin(2x)=0}}} , {{{cot(2x)}}} does not exist/is not defined,
and the function {{{f(x)=2cos(x)/cot(2x)}}} is not defined.
As long as {{{sin(2x)<>0}}} ,
{{{f(x)}}} has the same values as {{{g(x)=2cos(x)*tan(2x)}}} ,
but if {{{sin(2x)=0}}} {{{f(x) does not exist, while {{{g(x)=0}}} .
The graphs of {{{f(x)}}} and {{{g(x)}}} look alike,
except that where {{{sin(2x)=0}}} the graph of {{{f(x)}}} has a hole.
 
When {{{x=pi/4}}} , {{{cos(2x)=cos(pi/2)=0}}} .
When {{{x=3(pi/4)}}} , {{{cos(2x)=cos(3pi/2)=0}}} .
As {{{cos(2x)}}} is a periodic function with period {{{pi}}} ,
the zeros repeat at {{{x=pi/4+pi=5pi/4}}} , {{{x=3pi/4+pi=7pi/4}}} , and so on.
  
When {{{x=0=0*(pi/4)}}} , {{{sin(2x)=sin(0)=0}}} .
When {{{x=pi/2=2pi/4}}} , {{{sin(2x)=sin(pi)=0}}} .
As {{{sin(2x)}}} is a periodic function with period {{{pi}}} ,
the zeros repeat at {{{x=0+pi=pi=4(pi/4)}}} , {{{x=pi/2+pi=3pi/2=6(pi/4)}}} , and so on.
 
At every multiple of {{{pi/4}}} ,
either {{{cos(2x)=0}}} or {{{sin(2x)=0}}} .
 
The restrictions can be summarized as {{{highlight(x<>k(pi/4))}}} for every integer {{{k}}} .
 
The graph of the function {{{y=2cos(x)/cot(2x)}}} looks like the one below,
with holes and vertical asymptotes drawn in black:
{{{drawing(1000,300,-1,9,-5,5,
graph(1000,300,-1,9,-5,5,2cos(x)*tan(2x)),
line(0.785,-6,0.785,6),line(2.356,-6,2.356,6),
line(3.927,-6,3.927,6),line(5.498,-6,5.498,6),
line(7.069,-6,7.069,6),line(8.639,-6,8.639,6),
line(-0.785,-6,-0.785,6),circle(0,0,0.04),
circle(1.571,0,0.04),circle(3.142,0,0.04),circle(4.712,0,0.04),
circle(6.283,0,0.04),circle(7.854,0,0.04)
)}}}