Giải các phương trình sau 8cos^4x - 4cos2x + sin4x - 4 = 0
8 cos4x – 4 cos 2x + sin 4x – 4 = 0
\[ \Leftrightarrow 8{\left( {\frac{{1 + \cos 2x}}{2}} \right)^2} - 4\cos 2x + \sin 4x - 4 = 0\]
\[ \Leftrightarrow 2(1 + 2\cos 2x + {\cos ^2}2x) - 4\cos 2x + \sin 4x - 4 = 0\]
\[ \Leftrightarrow 2{\cos ^2}2x + \sin 4x - 2 = 0\]
\[ \Leftrightarrow 1 + \cos 4x + \sin 4x - 2 = 0\]
\[ \Leftrightarrow \cos 4x + \sin 4x = 1\]
\[ \Leftrightarrow \frac{{\sqrt 2 }}{2}\cos 4x + \frac{{\sqrt 2 }}{2}\sin 4x = \frac{{\sqrt 2 }}{2}\]
\[ \Leftrightarrow \sin \frac{\pi }{4}\cos 4x + \cos \frac{\pi }{4}\sin 4x = \frac{{\sqrt 2 }}{2}\]
\[ \Leftrightarrow \sin \left( {4x + \frac{\pi }{4}} \right) = \sin \frac{\pi }{4}\]
\[ \Leftrightarrow \left[ \begin{array}{l}4x + \frac{\pi }{4} = \frac{\pi }{4} + k2\pi \\4x + \frac{\pi }{4} = \frac{{3\pi }}{4} + k2\pi \end{array} \right.\]
\[ \Leftrightarrow \left[ \begin{array}{l}4x + \frac{\pi }{4} = \frac{\pi }{4} + k2\pi \\4x + \frac{\pi }{4} = \frac{{3\pi }}{4} + k2\pi \end{array} \right.\]
\[ \Leftrightarrow \left[ \begin{array}{l}4x = k2\pi \\4x = \frac{\pi }{2} + k2\pi \end{array} \right.\]
\[ \Leftrightarrow \left[ \begin{array}{l}4x = k2\pi \\4x = \frac{\pi }{2} + k2\pi \end{array} \right.\]
\[ \Leftrightarrow \left[ \begin{array}{l}x = \frac{{k\pi }}{2}\\x = \frac{\pi }{8} + \frac{{k\pi }}{2}\end{array} \right.\,\,(k \in \mathbb{Z})\]
Vậy phương trình đã cho có các nghiệm là \[x = \frac{{k\pi }}{2}\]; \[x = \frac{\pi }{8} + \frac{{k\pi }}{2}\,\,\,(k \in \mathbb{Z})\]