PT $\Leftrightarrow 2\sqrt{3} \cos^2x+2\sin3x\cos x-\sin4x-\sqrt{3}=\sqrt{3}\sin x+\cos x $$\Leftrightarrow \sqrt{3} (\cos2x+1)+(\sin 4x +\sin 2x)-\sin4x-\sqrt{3}=\sqrt{3}\sin x+\cos x $$\Leftrightarrow \sqrt{3}\cos 2x+\sin 2x =\sqrt{3}\sin x+\cos x $$\Leftrightarrow \sin\left ( 2x+\frac{\pi}{3}\right ) = \sin\left ( x+\frac{\pi}{6}\right ) $$\Leftrightarrow \left[ {\begin{matrix}2x+\frac{\pi}{3}=x+\frac{\pi}{6}+k2\pi\\2x+\frac{\pi}{3}=\pi-x-\frac{\pi}{6}+k2\pi \end{matrix}} \right.$$\Leftrightarrow \left[
{\begin{matrix}x=-\frac{\pi}{6}+k2\pi\\x=\frac{\pi}{6}+k\frac{2\pi}{3}
\end{matrix}} \right. (k \in \mathbb{Z}).$Ta cần điều kiện $ \sin\left ( x+\frac{\pi}{6}\right ) \ne 0\Leftrightarrow x \ne -\frac{\pi}{6}+l\pi$.Vậy $x=\frac{\pi}{6}+k\frac{2\pi}{3}
, (k \in \mathbb{Z}).$
PT $\Leftrightarrow 2\sqrt{3} \cos^2x+2\sin3x\cos x-\sin4x-\sqrt{3}=\sqrt{3}\sin x+\cos x $$\Leftrightarrow \sqrt{3} (\cos2x+1)+(\sin 4x +\sin 2x)-\sin4x-\sqrt{3}=\sqrt{3}\sin x+\cos x $$\Leftrightarrow \sqrt{3}\cos 2x+\sin 2x =\sqrt{3}\sin x+\cos x $$\Leftrightarrow \sin\left ( 2x+\frac{\pi}{3}\right ) = \sin\left ( x+\frac{\pi}{6}\right ) $$\Leftrightarrow \left[ {\begin{matrix}2x+\frac{\pi}{3}=x+\frac{\pi}{6}+k2\pi\\2x+\frac{\pi}{3}=\pi-x-\frac{\pi}{6}+k2\pi \end{matrix}} \right.$$\Leftrightarrow \left[
{\begin{matrix}x=-\frac{\pi}{6}+k2\pi\\2x+\frac{\pi}{3}=\frac{\pi}{6}+k\frac{2\pi}{3}
\end{matrix}} \right. (k \in \mathbb{Z}).$
PT $\Leftrightarrow 2\sqrt{3} \cos^2x+2\sin3x\cos x-\sin4x-\sqrt{3}=\sqrt{3}\sin x+\cos x $$\Leftrightarrow \sqrt{3} (\cos2x+1)+(\sin 4x +\sin 2x)-\sin4x-\sqrt{3}=\sqrt{3}\sin x+\cos x $$\Leftrightarrow \sqrt{3}\cos 2x+\sin 2x =\sqrt{3}\sin x+\cos x $$\Leftrightarrow \sin\left ( 2x+\frac{\pi}{3}\right ) = \sin\left ( x+\frac{\pi}{6}\right ) $$\Leftrightarrow \left[ {\begin{matrix}2x+\frac{\pi}{3}=x+\frac{\pi}{6}+k2\pi\\2x+\frac{\pi}{3}=\pi-x-\frac{\pi}{6}+k2\pi \end{matrix}} \right.$$\Leftrightarrow \left[
{\begin{matrix}x=-\frac{\pi}{6}+k2\pi\\x=\frac{\pi}{6}+k\frac{2\pi}{3}
\end{matrix}} \right. (k \in \mathbb{Z}).$
Ta cần điều kiện $ \sin\left ( x+\frac{\pi}{6}\right ) \ne 0\Leftrightarrow x \ne -\frac{\pi}{6}+l\pi$.Vậy $x=\frac{\pi}{6}+k\frac{2\pi}{3}
, (k \in \mathbb{Z}).$