Phương trình $\frac{{\sin 3x}}{{\cos 2x}} + \frac{{\cos 3x}}{{\sin 2x}} = \frac{2}{{\sin 3x}}$ có nghiệm là:
A. $x = \frac{\pi }{8} + k\frac{\pi }{4}$.
B. $x = \frac{\pi }{6} + k\frac{\pi }{3}$.
C. $x = \frac{\pi }{3} + k\frac{\pi }{2}$.
D. $x = \frac{\pi }{4} + k\pi $.
A. $x = \frac{\pi }{8} + k\frac{\pi }{4}$.
B. $x = \frac{\pi }{6} + k\frac{\pi }{3}$.
C. $x = \frac{\pi }{3} + k\frac{\pi }{2}$.
D. $x = \frac{\pi }{4} + k\pi $.
Chọn B.
Điều kiện $\left\{ \begin{array}{l}\cos 2x \ne 0\\\sin 2x \ne 0\\\sin 3x \ne 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}2x \ne \frac{{k\pi }}{2}\\3x \ne k\pi \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ne \frac{{k\pi }}{4}\\x \ne \frac{{k\pi }}{3}\end{array} \right.$
$ \Leftrightarrow \frac{{\sin 3x.\sin 2x + \cos 2x.\cos 3x}}{{\sin 2x.\cos 2x}} = \frac{2}{{\sin 3x}}$
$ \Leftrightarrow \frac{{2{\mathop{\rm cosx}\nolimits} }}{{\sin 4x}} = \frac{2}{{\sin 3x}}$$ \Leftrightarrow \sin 3x.\cos x = \sin 4x$$ \Leftrightarrow \frac{1}{2}\left( {\sin 2x + {\mathop{\rm s}\nolimits} {\rm{in4}}x} \right) = \sin 4x$
$ \Leftrightarrow \sin 2x = \sin 4x$$ \Leftrightarrow \left[ \begin{array}{l}2x = 4x + k2\pi \\2x = \pi - 4x + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = - k\pi \\x = \frac{\pi }{6} + \frac{{k\pi }}{3}\end{array} \right.$
So sánh với điều kiện, ta nhận $x = \frac{\pi }{6} + k\frac{\pi }{3}$.
Điều kiện $\left\{ \begin{array}{l}\cos 2x \ne 0\\\sin 2x \ne 0\\\sin 3x \ne 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}2x \ne \frac{{k\pi }}{2}\\3x \ne k\pi \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ne \frac{{k\pi }}{4}\\x \ne \frac{{k\pi }}{3}\end{array} \right.$
$ \Leftrightarrow \frac{{\sin 3x.\sin 2x + \cos 2x.\cos 3x}}{{\sin 2x.\cos 2x}} = \frac{2}{{\sin 3x}}$
$ \Leftrightarrow \frac{{2{\mathop{\rm cosx}\nolimits} }}{{\sin 4x}} = \frac{2}{{\sin 3x}}$$ \Leftrightarrow \sin 3x.\cos x = \sin 4x$$ \Leftrightarrow \frac{1}{2}\left( {\sin 2x + {\mathop{\rm s}\nolimits} {\rm{in4}}x} \right) = \sin 4x$
$ \Leftrightarrow \sin 2x = \sin 4x$$ \Leftrightarrow \left[ \begin{array}{l}2x = 4x + k2\pi \\2x = \pi - 4x + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = - k\pi \\x = \frac{\pi }{6} + \frac{{k\pi }}{3}\end{array} \right.$
So sánh với điều kiện, ta nhận $x = \frac{\pi }{6} + k\frac{\pi }{3}$.