Xác định $a$ để hai phương trình :
       $2 \cos x . \cos 2x = 1+\cos 2x +\cos 3x                       (1)$
       $4 \cos ^2 x - \cos 3x = a \cos x +(4-a)(1+\cos 2x)    (2)$
tương đương.
Phương trình $(1)   \Leftrightarrow \cos x + \cos 3x = 1+2 \cos ^2x-1+\cos 3x$
            $\Leftrightarrow \cos x (2 \cos x -1)=0   \Leftrightarrow  \left[ \begin{array}{l}
\cos x = 0\\
\cos x = \frac{1}{2}
\end{array} \right.$
Phương trình $(2)  \Leftrightarrow  4 \cos ^2 x -(4 \cos ^3 x - 3 \cos x )= a \cos x +2(4-a) \cos^2 x $
            $\Leftrightarrow 4 \cos ^3 x +(4 -2a) \cos^2 x +(a-3) \cos x =0$
            $\Leftrightarrow \cos x (2 \cos x -1)[2 \cos x -(a-3)]=0$
            $\Leftrightarrow \left[ \begin{array}{l}
\cos x = 0\\
\cos x = \frac{1}{2}\\
\cos x = \frac{{a - 3}}{2}
\end{array} \right.$
Hai phương trình $(1)$ và $(2)$ tương đương với nhau khi
Hoặc a)  $\frac{a-3 }{2 }=0  \Leftrightarrow  a=3$
Hoặc b)  $\frac{a-3 }{ 2} =\frac{ 1}{ 2} \Leftrightarrow a=4$
Hoặc c)   $ \frac{ a-3}{ 2} >1   \Leftrightarrow a>5$
Hoặc d) $\Leftrightarrow \frac{ a-3}{ 2} <-1 \Leftrightarrow a<1$
(  Hai phương trình tương đương được với nhau khi phương trình $(2)$ chỉ có nghiệm của $\cos x =0 , \cos x =\frac{1 }{ 2} $  )
Đáp số :  $a=3  ;  a =4  ;  a<1  ;  a>5$

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