a)    Giải phương trình lượng giác    $3\cos x+\cos2x-\cos3x+1=2\sin x\sin2x$
b)    Tìm tất cả các giá trị của tham số $m$ để phương trình trên tương đương với phương trình sau: $m\cos3x+(4-8m)\sin^2x+(7m-4)\cos x+(8m-4)=0$
a) Áp dụng các công thức :
$c{\rm{os}}2{\rm{x = co}}{{\rm{s}}^2}x - {\sin ^2}x$
$c{\rm{os}}3{\rm{x}} = 4c{\rm{o}}{{\rm{s}}^3}x - 3c{\rm{osx}}$
$1 = {\mathop{\rm s}\nolimits} {\rm{i}}{{\rm{n}}^2}{\rm{x}} + c{\rm{o}}{{\rm{s}}^2}{\rm{x, sin2x}} = 2\sin {\rm{x}}c{\rm{osx}}$
Ta có phương trình xuất phát tương đương với
    $c{\rm{osx}}\left( {c{\rm{osx}} + 1} \right) = 0$                            $(1)$
$ \Leftrightarrow \left[ \begin{array}{l}
c{\rm{osx}} = 0\\
c{\rm{osx}} + 1 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \frac{\pi }{2} + k\pi \\
x = \left( {2k + 1} \right)\pi
\end{array} \right.\left( {k \in Z} \right)$

b) Đặt $t = c{\rm{osx}},\left| t \right| \le 1$, khi đó $(1)$ trở thành
    $t\left( {t + 1} \right) = 0$      $(1’)$   có nghiệm ${t_1} = 0,{t_2} = - 1$
Áp dụng các công thức:
$c{\rm{os}}3{\rm{x}} = 4c{\rm{o}}{{\rm{s}}^3}x - 3c{\rm{osx}}$
${\sin ^2}x = 1 - c{\rm{o}}{{\rm{s}}^2}x$
Phương trình xuất phát trở thành:
$c{\rm{osx}}\left( {c{\rm{osx + 1}}} \right)\left( {mc{\rm{osx}} + m - 1} \right) = 0$            $(2)$
$ \Leftrightarrow t\left( {t + 1} \right)\left( {mt + m - 1} \right) = 0,\left| t \right| \le 1$            $( 2’)$
$\left( 1 \right) \Leftrightarrow \left( 2 \right)$ khi và chỉ khi $m < \frac{1}{2},m = 1$
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