As long as [tex]\sin x\neq0[/tex], we have
[tex]\sin x\cot x-\cos^2x=\sin x\dfrac{\cos x}{\sin x}-\cos^2x=\cos x-\cos^2x=0[/tex]
From here, you can pull out one factor of [tex]\cos x[/tex]:
[tex]\cos x(1-\cos x)=0[/tex]
Then either [tex]\cos x=0[/tex] or [tex]\cos x=1[/tex].
In the first case, [tex]\cos x=0[/tex] whenever [tex]x=\pm\dfrac\pi2,\pm\dfrac{3\pi}2,\pm\dfrac{5\pi}2,\ldots[/tex], or every odd multiple of [tex]\dfrac\pi2[/tex]. We denote this by [tex]x=\dfrac{(2n+1)\pi}2[/tex] for [tex]n\in\mathbb Z[/tex] ([tex]n[/tex] is any integer).
In the second case, [tex]\cos x=1[/tex] whenever [tex]x=0,\pm2\pi,\pm4\pi,\ldots[/tex], or every even multiple of [tex]\pi[/tex]. We write this as [tex]x=2n\pi[/tex] for [tex]n\in\mathbb Z[/tex].
I'm not sure what you mean by "principal values", but I'd guess it refers to any solutions to the equation for [tex]0\le x<2\pi[/tex]. In that case, you'd have only 3 solutions, [tex]x=\dfrac\pi2,\dfrac{3\pi}2,0[/tex].
However, we have to throw out the solution [tex]x=0[/tex], because that makes the left hand side of the original equation undefined. So the general solution is actually [tex]x=\dfrac{(2n+1)\pi}2[/tex] for [tex]n\in\mathbb Z[/tex] and [tex]x=2n\pi[/tex] for [tex]n\in\mathbb Z\setminus\{0\}[/tex] (all non-zero integers).
