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Câu hỏi: Phương trình ${5 - 2\cos 2x - 8\sin x = 0}$ có nghiệm là
A. $\left[ \begin{matrix}
x=\dfrac{\pi }{6}+k2\pi \\
x=\dfrac{5\pi }{6}+k2\pi \\
\end{matrix} \right.\left( k\in \mathbb{Z} \right) $.
B. $ \left[ \begin{matrix}
x=\dfrac{\pi }{6}+k\pi \\
x=\dfrac{5\pi }{6}+k\pi \\
\end{matrix} \right.\left( k\in \mathbb{Z} \right)$.
C. $\left[ \begin{matrix}
x=\dfrac{\pi }{6}+k2\pi \\
x=-\dfrac{\pi }{6}+k2\pi \\
\end{matrix} \right.\left( k\in \mathbb{Z} \right) $.
D. $\left[ \begin{matrix}
x=\dfrac{\pi }{6}+k\pi \\
x=-\dfrac{\pi }{6}+k\pi \\
\end{matrix} \right.\left( k\in \mathbb{Z} \right)$
A. $\left[ \begin{matrix}
x=\dfrac{\pi }{6}+k2\pi \\
x=\dfrac{5\pi }{6}+k2\pi \\
\end{matrix} \right.\left( k\in \mathbb{Z} \right) $.
B. $ \left[ \begin{matrix}
x=\dfrac{\pi }{6}+k\pi \\
x=\dfrac{5\pi }{6}+k\pi \\
\end{matrix} \right.\left( k\in \mathbb{Z} \right)$.
C. $\left[ \begin{matrix}
x=\dfrac{\pi }{6}+k2\pi \\
x=-\dfrac{\pi }{6}+k2\pi \\
\end{matrix} \right.\left( k\in \mathbb{Z} \right) $.
D. $\left[ \begin{matrix}
x=\dfrac{\pi }{6}+k\pi \\
x=-\dfrac{\pi }{6}+k\pi \\
\end{matrix} \right.\left( k\in \mathbb{Z} \right)$
Ta có: $5-2\cos 2x-8\sin x=0\Leftrightarrow 5-2(1-2si{{n}^{2}}x)-8\sin x=0$
$\Leftrightarrow 4{{\sin }^{2}}x-8\sin x+3=0\Leftrightarrow \left[ \begin{aligned}
& \sin x=\dfrac{3}{2}\left( loai \right) \\
& \sin x=\dfrac{1}{5} \\
\end{aligned} \right.\Leftrightarrow \left[ \begin{aligned}
& x=\dfrac{\pi }{6}+k2\pi \\
& x=\dfrac{5\pi }{6}+k2\pi \\
\end{aligned} \right.\left( k\in \mathbb{Z} \right)$
$\Leftrightarrow 4{{\sin }^{2}}x-8\sin x+3=0\Leftrightarrow \left[ \begin{aligned}
& \sin x=\dfrac{3}{2}\left( loai \right) \\
& \sin x=\dfrac{1}{5} \\
\end{aligned} \right.\Leftrightarrow \left[ \begin{aligned}
& x=\dfrac{\pi }{6}+k2\pi \\
& x=\dfrac{5\pi }{6}+k2\pi \\
\end{aligned} \right.\left( k\in \mathbb{Z} \right)$
Đáp án A.