solve the equation on the interval $0,2\\pi)$. $\\sqrt{3}\\cot(x)-1=-2$ enter the smaller radian value first…

Explanation:

Step1: Isolate (\cot(x))

Add (1) to both sides of the equation (\sqrt{3}\cot(x)-1 = - 2): (\sqrt{3}\cot(x)=-2 + 1=-1). Then divide both sides by (\sqrt{3}): (\cot(x)=-\frac{1}{\sqrt{3}}). Since (\cot(x)=\frac{\cos(x)}{\sin(x)}), we know that (\tan(x)=-\sqrt{3}) (because (\tan(x)=\frac{1}{\cot(x)})).

Step2: Find (x) in ([0,2\pi))

We know that (\tan(x)=-\sqrt{3}). The general solution of (\tan(x)=a) is (x = n\pi+\arctan(a)), (n\in\mathbb{Z}). For (a =-\sqrt{3}), (\arctan(-\sqrt{3})=-\frac{\pi}{3}). When (n = 1): (x=\pi-\frac{\pi}{3}=\frac{2\pi}{3}). When (n = 2): (x = 2\pi-\frac{\pi}{3}=\frac{5\pi}{3}).

Answer:

(x=\frac{2\pi}{3},\frac{5\pi}{3})