Xác định tham số $m$ để phương trình: $\cos3x-\cos2x+m\cos x-1=0$
Có đúng bảy nghiệm khác nhau thuộc khoảng $ - \frac{\pi }{2} < x < 2\pi $
Do $c{\rm{os}}3{\rm{x}} = 4c{\rm{o}}{{\rm{s}}^3}x - 3c{\rm{osx}},1 + c{\rm{os}}2{\rm{x}} = 2c{\rm{o}}{{\rm{s}}^2}x$ nên phương trình đã cho tương đương với:
    $4c{\rm{o}}{{\rm{s}}^3}x - 3c{\rm{osx}} - 2c{\rm{o}}{{\rm{s}}^2}x + mc{\rm{osx}} = 0$
$ \Leftrightarrow c{\rm{osx}}\left( {4c{\rm{o}}{{\rm{s}}^2}x - 2c{\rm{osx}} + m - 3} \right) = 0$
Với $ - \frac{\pi }{2} < x < 2\pi $ phương trình $c{\rm{osx = 0}}$ có hai nghiệm ${x_1} = \frac{\pi }{2}$, ${x_2} = \frac{{3\pi }}{2}$
$ \Rightarrow $cần chọn m để phương trình $4c{\rm{o}}{{\rm{s}}^2}x - 2c{\rm{osx}} + m - 3 = 0$ có đúng $5$ nghiệm thuộc khoảng $\left( { - \frac{\pi }{2};2\pi } \right)$ và khác ${x_1},{x_2}$.
Đặt $t = \cos {\rm{x}}$, khi đó cần chọn m để phương trình $f\left( t \right) = 4{t^2} - 2t + m - 3 = 0$ có hai nghiệm ${t_1},{t_2}$ với $ - 1 < {t_1} < 0 < {t_2} < 1$
$ \Leftrightarrow $$\left\{ \begin{array}{l}
f\left( { - 1} \right)f\left( 0 \right) < 0\\
f\left( 0 \right)f\left( 1 \right) < 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\left( {m + 2} \right)\left( {m - 3} \right) < 0\\
\left( {m - 3} \right)\left( {m - 1} \right) < 0
\end{array} \right. \Leftrightarrow 1 < m < 3$

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