$\begin{array}{l}a) \, \cos8x-\cos6x-\cos2x+1=0\\ \Leftrightarrow 2\cos^24x - 1 -(2\cos4x.\cos2x) + 1= 0\\ \Leftrightarrow \cos^24x - \cos4x.\cos2x = 0\\ \Leftrightarrow \cos4x(\cos4x - \cos2x) =0\\ \Leftrightarrow \left[ \begin{array}{l}\cos4x =0\\\cos4x = \cos2x\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}4x = \dfrac{\pi}{2} + k\pi\\4x = 2x + k2\pi\\4x = - 2x + k2\pi\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = \dfrac{\pi}{8} + k\dfrac{\pi}{4}\\x =k\pi\\x = k\dfrac{\pi}{3}\end{array} \right.\quad (k \in \Bbb Z)\\ b) \, \sin^3x\cos x - \cos^3x\sin x = 0\\ \Leftrightarrow \sin x\cos x(\sin^2x -\cos^2x) = 0\\ \Leftrightarrow \sin2x.\cos2x = 0\\ \Leftrightarrow \sin4x = 0\\ \Leftrightarrow 4x = k\pi\\ \Leftrightarrow x = k\dfrac{\pi}{4}\quad (k \in \Bbb Z)\\ c)\,\sin^22x-\cos^28x=\sin\left(\dfrac{\pi}{2}+6x\right)\\ \Leftrightarrow \dfrac{1 - \cos4x - 1 - \cos16x}{2} = \cos6x\\ \Leftrightarrow \cos4x + \cos16x + 2\cos6x = 0\\ \Leftrightarrow 2\cos10x.\cos6x + 2 \cos6x = 0\\ \Leftrightarrow \cos6x(\cos10x + 1) = 0\\ \Leftrightarrow \left[\begin{array}{l}\cos6x = 0\\\cos10x = -1\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}6x = \dfrac{\pi}{2} + k\pi\\10x = \pi + k2\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{12} + k\dfrac{\pi}{6}\\x = \dfrac{\pi}{10} + k\dfrac{\pi}{5}\end{array}\right.\quad (k \in \Bbb Z)\\ \end{array}$
