1 câu trả lời
Đáp án:
\(\left[ \begin{array}{l} x = \frac{\pi }{2} + k2\pi \\ x = \frac{\pi }{4} + \frac{{m\pi }}{2} \end{array} \right.\,\,\left( {k,\,\,m \in Z} \right)\)
Giải thích các bước giải: \(\begin{array}{l} \,\,\,\,\,\,2{\sin ^3}x + \cos 2x = \sin x\\ \Leftrightarrow 2{\sin ^3}x + 1 - 2{\sin ^2}x - \sin x = 0\\ \Leftrightarrow 2{\sin ^3}x - 2{\sin ^2}x - \sin x + 1 = 0\\ \Leftrightarrow \left( {\sin x - 1} \right)\left( {2{{\sin }^2}x - 1} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} \sin x = 1\\ \sin x = \frac{1}{{\sqrt 2 }}\\ \sin x = - \frac{1}{{\sqrt 2 }} \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = \frac{\pi }{2} + k2\pi \\ x = \frac{\pi }{4} + m2\pi \\ x = \frac{{3\pi }}{4} + m2\pi \\ x = - \frac{\pi }{4} + l2\pi \\ x = \frac{{5\pi }}{4} + l2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = \frac{\pi }{2} + k2\pi \\ x = \frac{\pi }{4} + \frac{{m\pi }}{2} \end{array} \right.\,\,\left( {k,\,\,m \in Z} \right). \end{array}\)
