Find solutions to $\cot(x)+\csc(x)=\sqrt3$ in range $[0,2\pi]$ What is the best way to do the above? Are there any tricks I should be aware of.
I know how to simplify it to $\dfrac{\cos(x)}{\sin(x)} + \dfrac{1}{\sin(x)} = \sqrt{3}$
so multiplying both sides by $\sin(x)$, we get $\cos(x)+1=\sqrt{3}\sin(x)$.
But I'm stuck from there.

Let $x = 2t$, then we can start at where you left off: 
$\cos x + 1 = \sqrt{3}\sin x\Rightarrow \cos (2t)+1 = \sqrt{3}\sin (2t) \Rightarrow 2\cos^2 t = 2\sqrt{3}\sin t\cos t \Rightarrow 2\cos t(\cos t - \sqrt{3}\sin t) = 0\Rightarrow \cos t = 0 \Rightarrow t = (2n+1)\dfrac{\pi}{2} \Rightarrow x = 2t = (2n+1)\pi, n \in \mathbb{Z}$ or $\cos t - \sqrt{3}\sin t = 0 \Rightarrow 2\cos (t+\frac{\pi}{3})=0 \Rightarrow t+\dfrac{\pi}{3} = (2n+1)\dfrac{\pi}{2} \Rightarrow t = n\pi + \dfrac{\pi}{6} \Rightarrow x = 2t = 2n\pi + \dfrac{\pi}{3}, n \in \mathbb{Z}$. Let $n = 0$ we obtain $x = \pi, \dfrac{\pi}{3}$ are the only $2$ solutions that belong to the range $[0,2\pi]$, and they are the only solutions.
