Giải phương trình cos2x + 3sinx -2=0
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1 câu trả lời 4344
\(\begin{array}{l}
cos2x{\rm{ }} + {\rm{ }}3sinx{\rm{ }} - 2 = 0\\
\Leftrightarrow 1 - 2{\sin ^2}x + 3\sin x - 2 = 0\\
\Leftrightarrow 2{\sin ^2}x - 3\sin x + 1 = 0\\
\Leftrightarrow 2{\sin ^2}x - 2\sin x - \sin x + 1 = 0\\
\Leftrightarrow 2\sin x\left( {\sin x - 1} \right) - \left( {\sin x - 1} \right) = 0\\
\Leftrightarrow \left( {\sin x - 1} \right)\left( {2\sin x - 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sin x - 1 = 0\\
2\sin x - 1 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\sin x = 1\\
\sin x = \frac{1}{2}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \frac{\pi }{2} + k2\pi \\
x = \frac{\pi }{6} + k2\pi \\
x = \frac{{5\pi }}{6} + k2\pi
\end{array} \right.\left( {k \in Z} \right)\\
vay:S = \{ \frac{\pi }{2} + k2\pi ;\frac{\pi }{6} + k2\pi ;\frac{{5\pi }}{6} + k2\pi \}
\end{array}\)
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