Giải các phương trình sau
1) $\cos x- \cos2x+ \cos3x = \frac{1}{2}$            (1)
2) $\sin 4x - 4\sin x  - (\cos4x-4\cos x) = 1$ (2)
$1)$ Để nhân hai vế với $c{\rm{os}}\frac{x}{2}$ ta xét trường hợp $c{\rm{os}}\frac{x}{2} = 0$
    $ \Leftrightarrow x = \pi  + 2k\pi $. Khi đó dễ thấy (1) không thỏa mãn vì $c{\rm{osx}},c{\rm{os2x}},c{\rm{os}}3{\rm{x}}$ bằng $ \pm 1$ nên $VT\left( 1 \right) \ne \frac{1}{2}$.
Trường hợp $c{\rm{os}}\frac{x}{2} \ne 0 \Leftrightarrow x \ne \pi  + 2k\pi $: nhân 2 vế với $c{\rm{os}}\frac{x}{2}$ rồi biến tích thành tổng ta được
$(1) \Leftrightarrow \frac{1}{2}\left( {c{\rm{os}}\frac{{3{\rm{x}}}}{2} + c{\rm{os}}\frac{x}{2} - c{\rm{os}}\frac{{5{\rm{x}}}}{2} - c{\rm{os}}\frac{{3{\rm{x}}}}{2} + c{\rm{os}}\frac{{7{\rm{x}}}}{2} + c{\rm{os}}\frac{{5{\rm{x}}}}{2}} \right) = \frac{1}{2}c{\rm{os}}\frac{x}{2}$
$ \Leftrightarrow c{\rm{os}}\frac{{7{\rm{x}}}}{2} = 0 \Leftrightarrow x = \frac{\pi }{7} + 4n\frac{\pi }{7} = \frac{{4n + 1}}{7}\pi $
Để nghiệm này thỏa mãn điều kiện $x \ne \pi  + 2k\pi  = \left( {2k + 1} \right)\pi $ ta chỉ được lấy $n \in N$ sao cho $\frac{{4n + 1}}{7} \ne $ số lẻ

$2)   (2)$ $ \Leftrightarrow 4\left( {c{\rm{osx}} - {\mathop{\rm s}\nolimits} {\rm{inx}}} \right) = c{\rm{os}}4{\rm{x}} - 1 - \sin 4{\rm{x}}$
                                            $ = c{\rm{o}}{{\rm{s}}^2}2{\rm{x  -  si}}{{\rm{n}}^2}2{\rm{x}} + {\left( {c{\rm{os}}2{\rm{x}} - \sin 2{\rm{x}}} \right)^2}$
                                            $ = \left( {c{\rm{os}}2{\rm{x}} - \sin 2{\rm{x}}} \right).2c{\rm{os}}2{\rm{x}}$
                                            $ = 2\left( {c{\rm{os}}2{\rm{x}} - \sin 2{\rm{x}}} \right)\left( {c{\rm{o}}{{\rm{s}}^2}{\rm{x}} - {\mathop{\rm s}\nolimits} {\rm{i}}{{\rm{n}}^2}{\rm{x}}} \right)$
$ \Leftrightarrow \left( {c{\rm{osx}} - {\mathop{\rm s}\nolimits} {\rm{inx}}} \right)\left[ {2 - \left( {c{\rm{os}}2{\rm{x}} - \sin 2{\rm{x}}} \right)\left( {c{\rm{osx}} + {\mathop{\rm s}\nolimits} {\rm{inx}}} \right)} \right] = 0$
$ \Leftrightarrow \left( {c{\rm{osx}} - {\mathop{\rm s}\nolimits} {\rm{inx}}} \right).2\left[ {1 - c{\rm{os}}\left( {2{\rm{x}} + \frac{\pi }{4}} \right)c{\rm{os}}\left( {x - \frac{\pi }{4}} \right)} \right] = 0$
Khi đó:
Hoặc $\left( {c{\rm{osx}} - {\mathop{\rm s}\nolimits} {\rm{inx}}} \right) = 0 \Leftrightarrow x = \frac{\pi }{4} + k\pi $
Hoặc $c{\rm{os}}\left( {2{\rm{x}} + \frac{\pi }{4}} \right).c{\rm{os}}\left( {x - \frac{\pi }{4}} \right) = 1$
    $ \Leftrightarrow \frac{1}{2}\left[ {c{\rm{os}}3{\rm{x}} + c{\rm{os}}\left( {x + \frac{\pi }{2}} \right)} \right] = 1 \Leftrightarrow c{\rm{os}}3{\rm{x}} - {\mathop{\rm s}\nolimits} {\rm{inx}} = 2$
Dẫn tới hệ $\left\{ \begin{array}{l}
c{\rm{os}}3{\rm{x}} = 1\\
- {\mathop{\rm s}\nolimits} {\rm{inx}} = 1
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
4c{\rm{o}}{{\rm{s}}^3}x - 3c{\rm{osx}} = 1\;{\rm{   (3)}}\\
{\mathop{\rm s}\nolimits} {\rm{inx}} = - 1{\rm{                   (4)}}
\end{array} \right.$
Hệ này vô nghiệm do ${\mathop{\rm s}\nolimits} {\rm{inx}} = 1$ thì $c{\rm{os}} = 0$ nên  $(3)$ không thỏa mãn

Vậy nghiệm  là $x = \frac{\pi }{4} + k\pi $

Thẻ

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