Đáp án:
$a) \left[\begin{array}{l} x =\dfrac{\pi}{6}+ k \pi(k \in \mathbb{Z})\\x=\dfrac{\pi}{3}+ k \pi(k \in \mathbb{Z})\end{array} \right.\\ b)x=k 2 \pi(k \in \mathbb{Z})\\ c)x=\dfrac{\pi}{2}+k 2 \pi(k \in \mathbb{Z}).$
Giải thích các bước giải:
$a)\sqrt{3}\cot^2x-4\cot x +\sqrt{3}=0\\ \Leftrightarrow \sqrt{3}\cot^2x-\cot x -3\cot x+\sqrt{3}=0\\ \Leftrightarrow \cot x (\sqrt{3}\cot x-1) -\sqrt{3}(\sqrt{3}\cot x-\sqrt{3})=0\\ \Leftrightarrow (\cot x -\sqrt{3})(\sqrt{3}\cot x-1)=0\\ \Leftrightarrow \left[\begin{array}{l} \cot x -\sqrt{3}=0\\ \sqrt{3}\cot x-1=0\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} \cot x =\sqrt{3}\\ \cot x=\dfrac{1}{\sqrt{3}}\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} \cot x =\cot\left(\dfrac{\pi}{6}\right)\\ \cot x=\cot\left(\dfrac{\pi}{3}\right)\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} x =\dfrac{\pi}{6}+ k \pi(k \in \mathbb{Z})\\x=\dfrac{\pi}{3}+ k \pi(k \in \mathbb{Z})\end{array} \right.\\ b)\cos^2x-4\cos x+3=0\\ \Leftrightarrow \cos^2x-\cos x-3\cos x+3=0\\ \Leftrightarrow \cos x(\cos x-1)-3(\cos x-1)=0\\ \Leftrightarrow (\cos x-3)(\cos x-1)=0\\ \Leftrightarrow \left[\begin{array}{l} \cos x-3=0\\ \cos x-1=0\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} \cos x=3(\text{Vô nghiệm})\\ \cos x=1\end{array} \right.\\ \Leftrightarrow \cos x=1\\ \Leftrightarrow x=k 2 \pi(k \in \mathbb{Z})\\ c)-\sin^2x+2\sin x-1=0\\ \Leftrightarrow -(\sin x-1)^2=0\\ \Leftrightarrow \sin x-1=0\\ \Leftrightarrow \sin x=1\\ \Leftrightarrow x=\dfrac{\pi}{2}+k 2 \pi(k \in \mathbb{Z}).$
