Cos 2x- cos 8x +cos 4x=1

Đáp án:

$\left\{ \matrix{ x = k\pi \hfill \cr x \approx {1 \over 2}\arccos 0,57 + k\pi \hfill \cr} \right.(k\in\mathbb Z)$

Lời giải:

$\eqalign{ & \cos 2x - \cos 8x + \cos 4x = 1 \cr & \Leftrightarrow \cos 2x - \left( {2{{\cos }^2}4x - 1} \right) + \left( {2{{\cos }^2}2x - 1} \right) = 1 \cr & \Leftrightarrow \cos 2x - \left[ {2{{\left( {2{{\cos }^2}2x - 1} \right)}^2} - 1} \right] + 2{\cos ^2}2x - 1 - 1 = 0 \cr & \Leftrightarrow \cos 2x - \left[ {2\left( {4{{\cos }^4}2x - 4{{\cos }^2}2x + 1} \right) - 1} \right] + 2{\cos ^2}2x - 2 = 0 \cr & \Leftrightarrow \cos 2x - 8{\cos ^4}2x + 8{\cos ^2}2x - 1 + 2{\cos ^2}2x - 2 = 0 \cr & \Leftrightarrow - 8{\cos ^4}2x + 10{\cos ^2}2x + \cos 2x - 3 = 0 \cr & \Leftrightarrow - 8{\cos ^4}2x + 8{\cos ^2}2x + 2{\cos ^2}2x + \cos 2x - 3 = 0 \cr & \Leftrightarrow - 8{\cos ^2}2x\left( {{{\cos }^2}2x - 1} \right) + \left( {\cos 2x - 1} \right)\left( {2\cos 2x + 3} \right) = 0 \cr & \Leftrightarrow - 8{\cos ^2}2x\left( {\cos 2x - 1} \right)\left( {\cos 2x + 1} \right) + \left( {\cos 2x - 1} \right)\left( {2\cos 2x + 3} \right) = 0 \cr & \Leftrightarrow \left( {\cos 2x - 1} \right)\left[ { - 8{{\cos }^3}2x - 8{{\cos }^2}2x + 2\cos 2x + 3} \right] = 0 \cr & \Leftrightarrow \left[ \matrix{ \cos 2x = 1 \hfill \cr - 8{\cos ^3}2x - 8{\cos ^2}2x + 2\cos 2x + 3 = 0 \hfill \cr} \right. \Leftrightarrow \left[ \matrix{ 2x = k2\pi \hfill \cr \cos 2x \approx 0,57 \hfill \cr} \right. \cr & \Leftrightarrow \left[ \matrix{ x = k\pi \hfill \cr x \approx {1 \over 2}\arccos 0,57 + k\pi \hfill \cr} \right.\,\,\left( {k \in Z} \right) \cr}$