## Algebra and Trigonometry 10th Edition

$x=\frac{\sqrt 2}{2}+n\pi$ and $x=arccot~5+n\pi$, where $n$ is an integer.
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.