Đáp án:
\[\left[ \begin{array}{l}
x = - \dfrac{\pi }{2} + k2\pi \\
x = - \dfrac{\pi }{6} + k2\pi \\
x = \dfrac{{7\pi }}{6} + k2\pi
\end{array} \right.\,\,\,\left( {k \in Z} \right)\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\cos 2x = 1 - 2{\sin ^2}x\\
\cos 2x - 3\sin x = 2\\
\Leftrightarrow \left( {1 - 2{{\sin }^2}x} \right) - 3\sin x - 2 = 0\\
\Leftrightarrow 1 - 2{\sin ^2}x - 3\sin x - 2 = 0\\
\Leftrightarrow - 2{\sin ^2}x - 3\sin x - 1 = 0\\
\Leftrightarrow 2{\sin ^2}x + 3\sin x + 1 = 0\\
\Leftrightarrow \left( {\sin x + 1} \right)\left( {2\sin x + 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sin x + 1 = 0\\
2\sin x + 1 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\sin x = - 1\\
\sin x = - \dfrac{1}{2}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = - \dfrac{\pi }{2} + k2\pi \\
x = - \dfrac{\pi }{6} + k2\pi \\
x = \pi - \left( { - \dfrac{\pi }{6}} \right) + k2\pi
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = - \dfrac{\pi }{2} + k2\pi \\
x = - \dfrac{\pi }{6} + k2\pi \\
x = \dfrac{{7\pi }}{6} + k2\pi
\end{array} \right.\,\,\,\left( {k \in Z} \right)
\end{array}\)
