Đáp án:

Giải thích các bước giải: \(\begin{array}{l} 3 - 4{\sin ^2}3x = 2\sin 3x\left( {2\cos 3x - 1} \right)\\ \Leftrightarrow 3 - \left( {4 - 4{{\cos }^2}3x} \right) = 2\sin 3x\left( {2\cos 3x - 1} \right)\\ \Leftrightarrow 4{\cos ^2}3x - 1 - 2\sin 3x\left( {2\cos 3x - 1} \right) = 0\\ \Leftrightarrow \left( {2\cos 3x - 1} \right)\left( {2\cos 3x + 1 - 2\sin 3x} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} 2\cos 3x - 1 = 0\\ \cos 3x - \sin 3x = - 1 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} \cos 3x = \frac{1}{2}\\ \cos \left( {3x + \frac{\pi }{4}} \right) = \frac{{ - 1}}{{\sqrt 2 }} \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x = \pm \frac{\pi }{9} + \frac{{k2\pi }}{3}\\ 3x + \frac{\pi }{4} = \pm \frac{{3\pi }}{4} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = \pm \frac{\pi }{9} + \frac{{k2\pi }}{3}\\ x = \frac{\pi }{6} + \frac{{k2\pi }}{3}\\ x = - \frac{\pi }{3} + \frac{{k2\pi }}{3} \end{array} \right. \end{array}\)

3 − 4sin²3x = 2sin3x(2cos3x − 1)

⇔ 3−(4 − 4cos²3x) = 2sin3x(2cos3x − 1)

⇔ 4cos²3x − 1 − 2sin3x(2cos3x − 1)=0

⇔ (2cos3x − 1)(2cos3x + 1 − 2sin3x)=0

⇔ \(\left[ \begin{array}{l}x=2\\x=-1\end{array} \right.\)