Giải phương trình : $4\sin3x.\cos2x=1+2\sin3x$
Đặt $\sin x = t,  |t| \le 1$. Ta biết rằng $\sin 3x = 3\sin x - 4\sin ^3 x=3t-4t^3, \cos 2x = 1-2\sin^2 x=1-2t^2$.
PT $\Leftrightarrow 4(3t-4t^3)(1-2t^2)=1+2(3t-4t^3)$ 
$\Leftrightarrow 32t^5-32t^3+6t-1=0$ 
$\Leftrightarrow (4t^2+2t-1)(8t^3-4t^2-4t+1)=0$ 
$\Leftrightarrow \left[ {\begin{matrix} t=\frac{1}{4}\left ( -1 \pm \sqrt5 \right ) \Rightarrow \left[ {\begin{matrix} x= \arcsin\frac{1}{4}\left ( -1 \pm \sqrt5 \right )+k2\pi \\ x= \pi -\arcsin\frac{1}{4}\left ( -1 \pm \sqrt5 \right )+k2\pi  \end{matrix}} \right.\\ 8t^3-4t^2-4t+1=0   (*)\end{matrix}} \right.$ 
Với PT $(*)$  ta đặt $t= y + \frac{1}{6}$ và thay vào $(*)$ ta được $y^3-\frac{7}{12}y=-\frac{7}{216}$.
Lại đặt $y=\frac{\sqrt 7}{3}z \implies 4z^3-3z=-\frac{1}{2\sqrt 7}$ 
Đặt $z = \cos \alpha \implies \cos 3 \alpha =-\frac{1}{2\sqrt 7} $ . Từ đây dùng phép thay thế ngược ta có
$\sin x=t=\frac{\sqrt 7}{3}z+ \frac{1}{6}=\frac{\sqrt 7}{3}\arccos-\frac{1}{6\sqrt 7}+ \frac{1}{6}$ 
và từ đây hoàn thành bài toán. 
thanks bác nhiều nhiều nhé –  dieuchinhthao 03-10-12 09:33 PM

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