Giải phương trình : $2(\cot x- \cos x)-3(\tan x-\sin x)=1 $
Điều kiện : $\sin 2x \ne 0.$
Đặt $t= \tan \frac{x}{2}$ thì PT đã cho đương với
      $2\left ( \frac{1-t^2}{2t}-\frac{1-t^2}{1+t^2} \right )-3\left ( \frac{2t}{1-t^2}-\frac{2t}{1+t^2} \right )=1$
$\Leftrightarrow t^6-t^5-13t^4+4t^3-t^2-3t+1=0$
$\Leftrightarrow (t^2+3t-1)(t^4-4t^3-1)=0$
$\Leftrightarrow \left[ {\begin{matrix} t=\frac{1}{2}\left (-3 \pm \sqrt{13} \right )\Rightarrow x = 2k\pi +2\arctan\frac{1}{2}\left (-3 \pm \sqrt{13} \right )\\t^4-4t^3-1=0        (*)\end{matrix}} \right.$
PT $(*) \Leftrightarrow t^4-4t^3+4t^2=4t^2+1$
             $\Leftrightarrow \left[ {t(t-2)} \right]^2=4t^2+1$
             $\Leftrightarrow \left[ {t(t-2)} \right]^2-2t(t-2)+1=2t^2+4t+2$
             $\Leftrightarrow \left[ {t(t-2)-1} \right]^2=2(t+1)^2$
             $\Leftrightarrow\left[ {\begin{matrix}  t^2-2t-1-\sqrt 2 (t+1)=0\\  t^2-2t-1+\sqrt 2 (t+1)=0    \text {(vô nghiệm)}\end{matrix}}    \right.$   
             $\Leftrightarrow\left[ {\begin{matrix}  t=1+\frac{1}{\sqrt 2}+\sqrt{\frac{5}{2}+2\sqrt 2}\\ t=1+\frac{1}{\sqrt 2}-\sqrt{\frac{5}{2}+2\sqrt 2} \end{matrix}} \right.$     
 Đến đây bạn tự viết nốt nghiệm nhé.
oh thanks bạn nhá –  magiamsinh16 24-09-12 11:26 PM
Nếu bạn thấy lời giải này chính xác và có ích đối với bạn. Hãy ấn vào mũi tên bên cạnh đáp án để bình chọn( vote up ) nhé! –  Trần Nhật Tân 24-09-12 07:35 AM

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