cos8x + sin^3 x. cosx – cos^3 x.sinx – 1 =0 cos^6 2x + sin^6 2x = 15/8 cos4x – 1/2 Giải các phương trình trên

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

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