Giải phương trình :  $3 \cos ^4 x - 4 \cos ^2 x  \sin ^2 x +\sin ^4 x =0     (1)$
Phương trình $(1)$ là phương trình có vế trái đẳng cấp bậc $4$  đối với $ \sin x $ và $\cos x .$
Nếu  $\cos x =0$   thì phương trình  $(1)$  có dạng  $1=0$  vậy đẳng thức sai nên $(1)$ không có nghiệm của   $\cos x=0$
Nếu $\cos x \neq  0$. Chia $2$  vế của $(1)$ cho $\cos ^4 x \neq  0     (1)$
        $\Leftrightarrow \tan ^4 x - 4 \tan ^2 x +3=0$
        $\left[ \begin{array}{l}
{\tan ^2}x = 1\\
{\tan ^2}x = 3
\end{array} \right.    \Leftrightarrow    \left[ \begin{array}{l}
{\mathop{\rm t}\nolimits} {\rm{anx}} = 1 = \tan \frac{\pi }{4}\\
{\mathop{\rm t}\nolimits} {\rm{anx}} = - 1 = \tan  - \frac{\pi }{4}\\
{\mathop{\rm t}\nolimits} {\rm{anx}} = \sqrt 3  = \tan \frac{\pi }{3}\\
{\mathop{\rm t}\nolimits} {\rm{anx}} = - \sqrt 3  = \tan \left( { - \frac{\pi }{3}} \right)
\end{array} \right.  \Leftrightarrow   \left[ \begin{array}{l}
x =  \pm \frac{\pi }{4} + k\pi \\
x =  \pm \frac{\pi }{3} + k\pi
\end{array} \right.$ 

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