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's exploit the half-angle identity  

$$\csc x+\cot x=\cot (x/2)$$

Then, we have $\cot(x/2) =\sqrt{3}$ or $\tan(x/2)=\sqrt{3}/3$.

Solutions are

$$x=2\arctan(\sqrt{3}/3)+2n\pi =\pi/3+2n\pi$$
Rearranging, we obtain
$$\sqrt3 \sin(x) - \cos(x) = 1 \implies \dfrac{\sqrt3}2 \sin(x) - \dfrac12 \cos(x) = \dfrac12 \implies \sin\left(x-\dfrac{\pi}6\right) = \sin\left(\dfrac{\pi}6\right)$$
This gives us that
$$x-\dfrac{\pi}6 = n\pi + (-1)^n \dfrac{\pi}6 \implies x = n\pi +(-1)^n \dfrac{\pi}6 + \dfrac{\pi}6$$
Further, $\sin(x)\neq 0 \implies x =2n\pi + \pi/3$.