Câu hỏi 12

\(\begin{array}{l}\,\,\,\,\,\,\,{\sin ^2}3x - {\cos ^2}4x = {\sin ^2}5x - {\cos ^2}6x.\\ \Leftrightarrow \frac{{1 - \cos 6x}}{2} - \frac{{1 + \cos 8x}}{2} = \frac{{1 - \cos 10x}}{2} - \frac{{1 + \cos 12x}}{2}\\ \Leftrightarrow \cos 6x + \cos 8x = \cos 10x + \cos 12x\\ \Leftrightarrow 2\cos 7x.\cos x = 2\cos 11x + \cos 12x\\ \Leftrightarrow 2\cos x\left( {\cos 7x - \cos 11x} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}\cos x = 0\\\cos 7x = \cos 11x\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{2} + k2\pi \\7x = 11x + k2\pi \\7x =  - 11x + k2\pi \end{array} \right.\, \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{2} + k2\pi \\x = \frac{{k\pi }}{2}\\x = \frac{{k\pi }}{9}\end{array} \right.\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)                                                                                                                                    Vậy \(S = \left\{ {\frac{{k\pi }}{2};\frac{{k\pi }}{9},\,\,k \in \mathbb{Z}} \right\}\).

Chọn D.