ĐKXĐ: \(sin2x\ne0\Leftrightarrow x\ne\dfrac{k\pi}{2}\)
\(\dfrac{sinx}{cosx}-\dfrac{3cosx}{sinx}=4\left(sinx+\sqrt{3}cosx\right)\)
\(\Leftrightarrow\dfrac{sin^2x-3cos^2x}{sin2x}=2\left(sinx+\sqrt{3}cosx\right)\)
\(\Leftrightarrow\dfrac{\left(sinx-\sqrt{3}cosx\right)\left(sinx+\sqrt{3}cosx\right)}{sin2x}=2\left(sinx+\sqrt{3}cosx\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx+\sqrt{3}cosx=0\\sinx-\sqrt{3}cosx=2sin2x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}sinx+\dfrac{\sqrt{3}}{2}cosx=0\\\dfrac{1}{2}sinx-\dfrac{\sqrt{3}}{2}cosx=sin2x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\dfrac{\pi}{3}\right)=0\\sin\left(x-\dfrac{\pi}{3}\right)=sin2x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{\pi}{3}=k\pi\\2x=x-\dfrac{\pi}{3}+k2\pi\\2x=\dfrac{4\pi}{3}-x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{3}+k\pi\\\\x=\dfrac{4\pi}{9}+\dfrac{k2\pi}{3}\end{matrix}\right.\) (đều thỏa mãn ĐKXĐ)
