1 câu trả lời
Đáp án:
Giải thích các bước giải: \[\begin{array}{l} 2{\cos ^2}x(\cos 2x - 4\sin x + 1) + 3{\sin ^2}x = 0\\ \Leftrightarrow 2{\cos ^2}x(2{\cos ^2}x - 4\sin x) + 3{\sin ^2}x = 0\\ \Leftrightarrow 4{\cos ^4}x - 8{\cos ^2}x.\sin x + 3{\sin ^2}x = 0\\ \Leftrightarrow 4{\cos ^4}x - 2{\cos ^2}x.\sin x - 6{\cos ^2}x.\sin x + 3{\sin ^2}x = 0\\ \Leftrightarrow 2{\cos ^2}x(2{\cos ^2}x - \sin x) - 3\sin x(2{\cos ^2}x - \sin x) = 0\\ \Leftrightarrow \left( {2{{\cos }^2}x - 3\sin x} \right)(2{\cos ^2}x - \sin x) = 0\\ \Leftrightarrow (2 - 2{\sin ^2}x - 3\sin x)(2 - 2{\sin ^2}x - \sin x) = 0\\ \Leftrightarrow \left[ \begin{array}{l} \sin x = \frac{1}{2} \Leftrightarrow \left[ \begin{array}{l} x = \frac{\pi }{6} + k2\pi \\ x = \frac{{5\pi }}{6} + k2\pi \end{array} \right.\\ \sin x = - 2 = > vn\\ \sin x = \frac{{ - 1 - \sqrt {17} }}{4} = > vn\\ \sin x = \frac{{ - 1 + \sqrt {17} }}{4} \Leftrightarrow \left[ \begin{array}{l} x = \arcsin \frac{{ - 1 + \sqrt {17} }}{4} + k2\pi \\ x = \pi - \arcsin \frac{{ - 1 + \sqrt {17} }}{4} + k2\pi \end{array} \right. \end{array} \right.\\ \end{array}\]
