Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
{\sin ^2}3x - {\cos ^2}4x = {\sin ^2}5x - {\cos ^2}6x\\
\Leftrightarrow {\cos ^2}6x - {\cos ^2}4x = {\sin ^2}5x - {\sin ^2}3x\\
\Leftrightarrow \left( {\cos 6x - \cos 4x} \right).\left( {\cos 6x + \cos 4x} \right) = \left( {\sin 5x - \sin 3x} \right).\left( {\sin 5x + \sin 3x} \right)\\
\Leftrightarrow - 2.\sin \dfrac{{6x + 4x}}{2}.\sin \dfrac{{6x - 4x}}{2}.2.\cos \dfrac{{6x + 4x}}{2}.\cos \dfrac{{6x - 4x}}{2} = 2.cos\dfrac{{5x + 3x}}{2}.\sin \dfrac{{5x - 3x}}{2}.2.sin\dfrac{{5x + 3x}}{2}.\cos \dfrac{{5x - 3x}}{2}\\
\Leftrightarrow - 4.\sin 5x.\sin x.\cos 5x.\cos x = 4.\cos 4x.\sin x.\sin 4x.\cos x\\
\Leftrightarrow - \sin 10x.\sin 2x = \sin 8x.\sin 2x\\
\Leftrightarrow \sin 2x.\left( {\sin 8x + \sin 10x} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sin 2x = 0\\
\sin 8x = - \sin 10x
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\sin 2x = 0\\
\sin 8x = \sin \left( { - 10x} \right)
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
2x = k\pi \\
8x = - 10x + k2\pi \\
8x = \pi + 10x + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{k\pi }}{2}\\
x = \dfrac{{k\pi }}{9}\\
x = - \dfrac{\pi }{2} + k\pi
\end{array} \right.\\
2,\\
\cos 3x - 4\cos 2x + 3\cos x - 4 = 0\\
\Leftrightarrow \left( {4{{\cos }^3}x - 3\cos x} \right) - 4.\left( {2{{\cos }^2}x - 1} \right) + 3\cos x - 4 = 0\\
\Leftrightarrow 4{\cos ^3}x - 8{\cos ^2}x = 0\\
\Leftrightarrow 4{\cos ^2}x\left( {\cos x - 2} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\cos x = 0\\
\cos x = 2\,\,\,\,\left( L \right)
\end{array} \right.\\
\Leftrightarrow \cos x = 0\\
\Leftrightarrow x = \dfrac{\pi }{2} + k\pi
\end{array}\)
