Đáp án:
\[B\]
Giải thích các bước giải:
ĐKXĐ: \(\left\{ \begin{array}{l}
\sin x \ne 0\\
\cos x \ne 0\\
\sin 2x \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\sin x \ne 0\\
\cos x \ne 0\\
2\sin x.\cos x \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\sin x \ne 0\\
\cos x \ne 0
\end{array} \right. \Leftrightarrow x \ne \dfrac{{k\pi }}{2}\)
Ta có:
\(\begin{array}{l}
\cot x - \tan x + 4\sin 2x = \dfrac{2}{{\sin 2x}}\\
\Leftrightarrow \dfrac{{\cos x}}{{\sin x}} - \dfrac{{\sin x}}{{\cos x}} + 4\sin 2x = \dfrac{2}{{\sin 2x}}\\
\Leftrightarrow \dfrac{{{{\cos }^2}x - {{\sin }^2}x}}{{\sin x.\cos x}} + 4\sin 2x = \dfrac{2}{{\sin 2x}}\\
\Leftrightarrow \dfrac{{\cos 2x}}{{\dfrac{1}{2}\sin 2x}} + 4\sin 2x = \dfrac{2}{{\sin 2x}}\\
\Leftrightarrow \dfrac{{2\cos 2x}}{{\sin 2x}} + 4\sin 2x = \dfrac{2}{{\sin 2x}}\\
\Leftrightarrow 2\cos 2x + 4{\sin ^2}2x = 2\\
\Leftrightarrow 2\cos 2x + 4.\left( {1 - {{\cos }^2}2x} \right) - 2 = 0\\
\Leftrightarrow 2\cos 2x + 4 - 4{\cos ^2}2x - 2 = 0\\
\Leftrightarrow - 4{\cos ^2}2x + 2\cos 2x + 2 = 0\\
\Leftrightarrow 2{\cos ^2}2x - \cos 2x - 1 = 0\\
\Leftrightarrow \left( {2{{\cos }^2}2x - 2\cos 2x} \right) + \left( {\cos 2x - 1} \right) = 0\\
\Leftrightarrow 2\cos 2x\left( {\cos 2x - 1} \right) + \left( {\cos 2x - 1} \right) = 0\\
\Leftrightarrow \left( {\cos 2x - 1} \right)\left( {2\cos 2x + 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\cos 2x - 1 = 0\\
2\cos 2x + 1 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\cos 2x = 1\\
\cos 2x = - \dfrac{1}{2}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
2x = k2\pi \\
2x = \pm \dfrac{{2\pi }}{3} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = k\pi \\
x = \pm \dfrac{\pi }{3} + k\pi
\end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)\\
x \in \left( {0;2\pi } \right) \Rightarrow x \in \left\{ {\pi ;\dfrac{\pi }{3};\dfrac{{4\pi }}{3};\dfrac{{2\pi }}{3};\dfrac{{5\pi }}{3}} \right\}
\end{array}\)
Mà \(x \ne \dfrac{{k\pi }}{2} \Rightarrow x \in \left\{ {\dfrac{\pi }{3};\dfrac{{2\pi }}{3};\dfrac{{4\pi }}{3};\dfrac{{5\pi }}{3}} \right\}\)
Do đó, có \(4\) nghiệm nằm trong khoảng \(\left( {0;2\pi } \right)\) thoả mãn
Vậy đáp án đúng là \(B\)
