Answer
$x=\frac{\sqrt 2}{2}+n\pi$ and $x=arccot~5+n\pi$, where $n$ is an integer.
Work Step by Step
Use: $-6~cot~x=-5~cot~x-cot~x$
$cot^2x-6~cot~x+5=0$
$cot^2x-5~cot~x-cot~x+5=0$
$cot~x(cot~x-5)-1(cot~x-5)=0$
$(cot~x-1)(cot~x-5)=0$
$cot~x-1=0$
$cot~x=1$
$x=arccot~1=\frac{\sqrt 2}{2}$
$cot~x-5=0$
$cot~x=5$
$x=arccot~5$
The period of $cot~x$ is $\pi$. So, add multiples of $\pi$ to each solution to find the general solution:
$x=\frac{\sqrt 2}{2}+n\pi$ and $x=arccot~5+n\pi$, where $n$ is an integer.
