Đáp án:
\({x_{\min }} = \frac{\pi }{6}\)
Giải thích các bước giải:
\[\begin{array}{l}
\left( {2\sin x - \cos x} \right)\left( {1 + \cos x} \right) = {\sin ^2}x\\
\Leftrightarrow \left( {2\sin x - \cos x} \right)\left( {1 + \cos x} \right) = \left( {1 - {{\cos }^2}x} \right)\\
\Leftrightarrow \left( {2\sin x - \cos x} \right)\left( {1 + \cos x} \right) = \left( {1 - \cos x} \right)\left( {1 + \cos x} \right)\\
\Leftrightarrow \left( {1 + \cos x} \right)\left( {2\sin x - \cos x - 1 + \cos x} \right) = 0\\
\Leftrightarrow \left( {1 + \cos x} \right)\left( {2\sin x - 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
1 + \cos x = 0\\
2\sin x - 1 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\cos x = - 1\\
\sin x = \frac{1}{2}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \pi + k2\pi \\
x = \frac{\pi }{6} + m2\pi \\
x = \frac{{5\pi }}{6} + l2\pi
\end{array} \right.\,\,\,\,\left( {k,\,\,m,\,\,l \in Z} \right).\\
Nghiem\,\,duong\,\,nho\,\,\,nhat \Leftrightarrow \left[ \begin{array}{l}
\pi + k2\pi \ge 0\\
\frac{\pi }{6} + m2\pi \ge 0\\
\frac{{5\pi }}{6} + l2\pi \ge 0
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
{x_{\min 1}} = \pi \\
{x_{\min 2}} = \frac{\pi }{6}\\
{x_{\min 3}} = \frac{{5\pi }}{6}
\end{array} \right. \Rightarrow {x_{\min }} = \frac{\pi }{6}.
\end{array}\]
