Giải phương trình: cos 4x + cos 6x + cos 2x =  - 1/2

Trả lời

Lời giải

Ta có:

\(\cos 4x + \cos 6x + \cos 2x = - \frac{1}{2}\)

\( \Leftrightarrow \cos 4x + 2\cos 4x\cos 2x = - \frac{1}{2}\)

\( \Leftrightarrow {\cos ^2}2x - 1 + 2\left( {{{\cos }^2}2x - 1} \right)\cos 2x = - \frac{1}{2}\)

\( \Leftrightarrow {\cos ^2}2x - 1 + 2{\cos ^4}2x - 2{\cos ^2}2x = - \frac{1}{2}\)

\( \Leftrightarrow 4{\cos ^4}2x - 2{\cos ^2}2x - 1 = 0\)

\( \Leftrightarrow \left[ \begin{array}{l}{\cos ^2}2x = \frac{{1 + \sqrt 5 }}{4}\\{\cos ^2}2x = \frac{{1 - \sqrt 5 }}{4}\left( {loai} \right)\end{array} \right.\)\( \Leftrightarrow \cos 2x = \pm \frac{{\sqrt {1 + \sqrt 5 } }}{2}\)

\( \Leftrightarrow \left[ \begin{array}{l}2x = \pm \arccos \frac{{\sqrt {1 + \sqrt 5 } }}{2} + k2\pi \\2x = \pm \arccos \left( { - \frac{{\sqrt {1 + \sqrt 5 } }}{2}} \right) + k2\pi \end{array} \right.\left( {k \in \mathbb{Z}} \right)\)

\( \Leftrightarrow \left[ \begin{array}{l}x = \pm \frac{1}{2}\arccos \frac{{\sqrt {1 + \sqrt 5 } }}{2} + k2\pi \\x = \pm \frac{1}{2}\arccos \left( { - \frac{{\sqrt {1 + \sqrt 5 } }}{2}} \right) + k2\pi \end{array} \right.\left( {k \in \mathbb{Z}} \right)\).