2sin^2 3x-2sin3x-4=0

$$\eqalign{ & 2{\sin ^2}3x - 2\sin 3x - 4 = 0 \cr & \Leftrightarrow \left[ \matrix{ \sin 3x = 2\,\,\left( {loai} \right) \hfill \cr \sin 3x = - 1 \hfill \cr} \right. \Leftrightarrow 3x = - {\pi \over 2} + k2\pi \cr & \Leftrightarrow x = - {\pi \over 6} + {{k2\pi } \over 3}\,\,\left( {k \in Z} \right) \cr} $$

2sin ²3x-2sin3x-4=0

⇔sin3x=2(loại) hoặc sin3x=-1

⇔3x=- $\pi$ /2 + k2 $\pi$

⇔x=- $\pi$ /6 + k2 $\pi$ /3