A nice identity can be used!

$\cot x-\tan2x = 0 \\ \frac{\cos x}{\sin x}-\frac{\sin2x}{\cos2x} = 0 \\ \cos2x\cos x-\sin2x\sin x = 0 \\ \cos\left(2x+x\right) = 0 \\ \cos3x = 0 \\ 3x = 90\text{˚}+180\text{˚}k\mbox{, for }k\in\mathbb{Z}$

which gives $x=30\text{˚}\mbox{, }90\text{˚}\mbox{, }150\text{˚}\mbox{, }210\text{˚}\mbox{, }270\text{˚}\mbox{, }330\text{˚}$ , for $0\leq x\leq360^{\mbox{o}}$ .

$\begin{aligned} \cot{x} - \tan{2x} & = 0 \\ \cot{x} & = \tan{2x} \\ \tan{(90^\circ - x)} & = \tan{2x} \\ \tan{(180\color{#3D99F6}{k}^\circ+90^\circ - x)} & = \tan{(2x)} \quad \quad \quad \quad \color{#3D99F6}{[\text{for } k = 0, 1, 2, 3, 4, 5]} \\ \Rightarrow 3x & = (180k+90)^\circ \\ x & = (2k+1)30^\circ \\ \Rightarrow \sum_{k=0}^5 x & = \sum_{k=0}^5 (2k+1)30^\circ \\ & = \left(\frac{2(5)(6)}{2} + 6 \right)30^\circ \\ & = (30+6)30^\circ \\ & = \boxed{1080}^\circ \end{aligned}$

$\begin{aligned}\tan \left(\frac{\pi}{2} - x\right) = \cot x &= \tan 2x\\ \Rightarrow \frac{\pi}{2} - x &= 2x + n\pi\\ \Rightarrow x &= (2n + 1)\frac{\pi}{6}\\ \end{aligned}$

$\therefore$ Solutions are $\dfrac{\pi}{6}, \dfrac{3\pi}{6}, \dfrac{5\pi}{6}, \dfrac{7\pi}{6}, \dfrac{9\pi}{6}, \dfrac{11\pi}{6}$

Hence, the sum $= \dfrac{36\pi}{6} = 36 \times 30 = \boxed{1080^{\circ}}$