\[\begin{array}{l}
a)\,\,\,\cos 8x + {\sin ^3}x.cosx - co{s^3}x.\sin x - 1 = 0\\
\Leftrightarrow \cos 8x + \sin x.\cos x\left( {{{\sin }^2}x - {{\cos }^2}x} \right) - 1 = 0\\
\Leftrightarrow \cos 8x + \frac{1}{2}\sin 2x.\left( { - \cos 2x} \right) - 1 = 0\\
\Leftrightarrow 1 - 2{\sin ^2}4x - \frac{1}{2}\sin 2x.\cos 2x - 1 = 0\\
\Leftrightarrow - 2{\sin ^2}4x - \frac{1}{4}\sin 4x = 0\\
\Leftrightarrow 8{\sin ^2}4x - \sin 4x = 0\\
\Leftrightarrow \sin 4x\left( {8\sin 4x - 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sin 4x = 0\\
\sin 4x = \frac{1}{8}
\end{array} \right..\\
b)\,\,\,{\cos ^6}2x + {\sin ^6}2x = \frac{{15}}{8}\cos 4x - \frac{1}{2}\\
\Leftrightarrow 1 - \frac{3}{4}{\sin ^2}2x = \frac{{15}}{8}\cos 4x - \frac{1}{2}\\
\Leftrightarrow \frac{3}{4}{\sin ^2}2x + \frac{{15}}{8}\cos 4x - \frac{3}{2} = 0\\
\Leftrightarrow \frac{3}{4}{\sin ^2}2x + \frac{{15}}{8}.\left( {1 - 2{{\sin }^2}2x} \right) - \frac{3}{2} = 0\\
\Leftrightarrow \frac{3}{4}{\sin ^2}2x + \frac{{15}}{8} - \frac{{15}}{4}{\sin ^2}2x - \frac{3}{2} = 0\\
\Leftrightarrow - 3{\sin ^2}2x = - \frac{3}{8} \Leftrightarrow {\sin ^2}2x = \frac{1}{8}\\
\Leftrightarrow \left[ \begin{array}{l}
\sin 2x = \frac{1}{{2\sqrt 2 }}\\
\sin 2x = - \frac{1}{{2\sqrt 2 }}
\end{array} \right..
\end{array}\]
