Em tách câu hỏi ra nhé
\(\begin{array}{l}
a)\,\cos x.\cos 7x = \cos 3x.\cos 5x\\
\Leftrightarrow \frac{1}{2}\left( {\cos 8x + \cos 6x} \right) = \frac{1}{2}\left( {\cos 8x + \cos 2x} \right)\\
\Leftrightarrow \cos 6x - \cos 2x = 0\\
\Leftrightarrow - 2\sin 4x.\sin 2x = 0\\
\Leftrightarrow \sin 4x.\sin 2x = 0 \Rightarrow D\\
2)\,{\sin ^4}x - {\cos ^4}x = 0\\
\Leftrightarrow \left( {{{\sin }^2}x - {{\cos }^2}x} \right)\left( {{{\sin }^2}x + {{\cos }^2}x} \right) = 0\\
\Leftrightarrow \cos 2x = 0\\
\Leftrightarrow x = \frac{\pi }{4} + \frac{{k\pi }}{2} \Rightarrow C\\
3)\,\sin 3x - 4\sin x\cos 2x = 0\\
\Leftrightarrow \sin 3x - 2\left( {\sin 3x - \sin x} \right) = 0\\
\Leftrightarrow 2\sin x - \sin 3x = 0\\
\Leftrightarrow 2\sin x - \left( {3\sin x - 4{{\sin }^3}x} \right) = 0\\
\Leftrightarrow 4{\sin ^3}x - \sin x = 0\\
\Leftrightarrow \sin x\left( {4{{\sin }^2}x - 1} \right) = 0\\
\Leftrightarrow \sin x\left( {1 - 2\cos 2x} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sin x = 0\\
\cos 2x = \frac{1}{2}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = k\pi \\
x = \pm \frac{\pi }{6} + n\pi
\end{array} \right. \Rightarrow B
\end{array}\)
