Đáp án đúng: A
Giải chi tiết:\(\begin{array}{l}\,\,\,\,\,\,\,2{\sin ^3}x - \cos 2x + \cos x = 0\\ \Leftrightarrow 2\left( {1 - {{\cos }^2}x} \right).\sin x + \left( { - 2{{\cos }^2}x + \cos x + 1} \right) = 0\\ \Leftrightarrow 2\left( {1 - \cos x} \right)\left( {1 + \cos x} \right).\sin x + \left( {1 - \cos x} \right)\left( {2\cos x + 1} \right) = 0\\ \Leftrightarrow \left( {1 - \cos x} \right)\left[ {2\sin x\left( {1 + \cos x} \right) + 2\cos x + 1} \right] = 0\\ \Leftrightarrow \left[ \begin{array}{l}\cos x = 1\,\,\,\left( 1 \right)\\2\sin x + 2\sin x\cos x + 2\cos x + 1 = 0\,\,\,\left( 2 \right)\end{array} \right.\\\left( 1 \right) \Leftrightarrow x = k2\pi \,\,\,\left( {k \in \mathbb{Z}} \right)\\\left( 2 \right) \Leftrightarrow 2\left( {\sin x + \cos x} \right) + 2\sin x\cos x + 1 = 0\end{array}\)
Đặt \(\sin x + \cos x = t\,\,\,\,\left( { - \sqrt 2 \le t \le \sqrt 2 } \right)\), khi đó ta có:
\(\begin{array}{l}{\left( {\sin x + cosx} \right)^2} = {t^2} \Leftrightarrow {\sin ^2}x + 2\sin x.\cos x + {\cos ^2}x = {t^2}\\ \Leftrightarrow 1 + 2\sin x.\cos x = {t^2} \Leftrightarrow \sin x.\cos x = \dfrac{{{t^2} - 1}}{2}\end{array}\) Phương trình (2) trở thành:
\(\begin{array}{l}2t + 2.\dfrac{{{t^2} - 1}}{2} + 1 = 0 \Leftrightarrow {t^2} + 2t = 0 \Leftrightarrow \left[ \begin{array}{l}t = 0\,\,\,\,\,\,\,\left( {tm} \right)\\t = - 2\,\,\left( {ktm} \right)\end{array} \right.\\ \Rightarrow \sin x + \cos x = 0 \Leftrightarrow \sqrt 2 \sin \left( {x + \dfrac{\pi }{4}} \right) = 0\\ \Leftrightarrow \sin \left( {x + \dfrac{\pi }{4}} \right) = 0 \Leftrightarrow x = \dfrac{{ - \pi }}{4} + k\pi \,\,\,\left( {k \in \mathbb{Z}} \right).\end{array}\)
Vậy \(S = \left\{ { - \dfrac{\pi }{4} + k\pi \,,\,\,k \in \mathbb{Z}} \right\}\).
