Đáp án:
Giải thích các bước giải:
\[\begin{array}{l}
2{\sin ^3}x - \cos 2x + cosx = 0\\
\Leftrightarrow 2sinx(1 - co{s^2}x) + (1 - 2{\cos ^2}x + \cos x) = 0\\
\Leftrightarrow 2\sin x(1 - \cos x)(1 + \cos x) - (2cosx + 1)(\cos x - 1) = 0\\
\Leftrightarrow (1 - \cos x)(2\sin x.\cos x + 2\sin x + 2\cos x + 1) = 0\\
\Leftrightarrow (1 - \cos x)({\sin ^2}x + {\cos ^2}x + 2\sin x\cos x + 2(\sin x + \cos x)) = 0\\
\Leftrightarrow (1 - \cos x)\left[ {{{(\sin x + \cos x)}^2} + 2(\sin x + \cos x)} \right] = 0\\
\Leftrightarrow (1 - \cos x)(\sin x + \cos x)(\sin x + \cos x + 2) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\cos x = 1\\
\sin x + \cos x = 0\\
\sin x + \cos x = - 2
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = k2\pi \\
\sin (x + \frac{\pi }{4}) = 0 \Leftrightarrow x = - \frac{\pi }{4} + k\pi \\
\sin (x + \frac{\pi }{4}) = - \sqrt 2 = > VN
\end{array} \right.
\end{array}\]
