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

PT $\Leftrightarrow \dfrac{1-\cos 6x-1-\cos 8x}{2}=\dfrac{1-\cos 10x - 1-\cos 12x }{2}$

$\Leftrightarrow \cos 6x+\cos 8x=\cos 10x+\cos 12x$

$\Leftrightarrow \cos 12x-\cos 6x+\cos 10x-\cos 8x=0$

$\Leftrightarrow -2\sin 9x\sin 3x -2\sin 9x\sin x=0$

$\Leftrightarrow -2\sin 9x(\sin 3x-\sin x)=0$

$\Leftrightarrow \sin 9x=0\Leftrightarrow x=\dfrac{k\pi}{9}$

hoặc $\sin 3x=\sin x\Leftrightarrow 3x=x+k2\pi; 3x=\pi-x+k2\pi$

$\Leftrightarrow x=k\pi; x=\dfrac{\pi}{4}+\dfrac{k\pi}{2}$