xuannambka

$ \Leftrightarrow 4\sin x\cos x - 3\left( {1 - 2{{\sin }^2}x} \right) + 6\sin x - 2\cos x - 7 = 0$
$ \Leftrightarrow 4\sin x\cos x + 6{\sin ^2}x + 6\sin x - 2c{\rm{os}}x - 10 = 0$
+ TH1: $\cos \left( {\frac{x}{2}} \right) = 0$ phương trình vô nghiệm.
+ TH2: $\cos \left( {\frac{x}{2}} \right) \ne 0$ đặt $t = \tan \left( {\frac{x}{2}} \right)$ta có phương trình
 
$4\left( {\frac{{2t}}{{1 + {t^2}}}} \right)\left( {\frac{{1 - {t^2}}}{{1 + {t^2}}}} \right) + 6{\left( {\frac{{2t}}{{1 + {t^2}}}} \right)^2} + 6\left( {\frac{{2t}}{{1 + {t^2}}}} \right) - 2\left( {\frac{{1 - {t^2}}}{{1 + {t^2}}}} \right) - 10 = 0$
 
$\Leftrightarrow \left[ {\begin{array}{*{20}{c}}
{t = \frac{3}{2} \Leftrightarrow x = 2\arctan \left( {\frac{3}{2}} \right) + k\pi }\\
{{t^3} + {t^2} + t - 1 = 0\left( 1 \right)}
\end{array}} \right.$
 
$\left( 1 \right) \Leftrightarrow {\left( {t + \frac{1}{3}} \right)^3} + \frac{2}{3}\left( {t + \frac{1}{3}} \right) - \frac{{24}}{{27}} = 0 \Leftrightarrow {y^3} + \frac{2}{3}y - \frac{{24}}{{27}} = 0$
 
$\begin{array}{l}
 \Leftrightarrow {\left( {v + \frac{\beta }{v}} \right)^3} + \frac{2}{3}\left( {v + \frac{\beta }{v}} \right) - \frac{{24}}{{27}} = 0\\
 \Leftrightarrow {v^6} + \left( {3\beta  + \frac{2}{3}} \right){v^4} - \frac{{34}}{{27}}{v^3} + \left( {3{\beta ^2} + \frac{2}{3}\beta } \right){v^2} + {\beta ^3} = 0
\end{array}$
 
Chọn $\beta  =  - \frac{2}{9}$ ta có phương trình: 
${z^2} - \frac{{34}}{{27}}z - \frac{8}{{729}} = 0 \Leftrightarrow \left[ {\begin{array}{*{20}{c}}
{z = \frac{{17 - 3\sqrt {33} }}{{27}}}\\
{z = \frac{{17 + 3\sqrt {33} }}{{27}}}
\end{array}} \right. \Leftrightarrow \left[ {\begin{array}{*{20}{c}}
{v = \frac{{\sqrt[3]{{17 - 3\sqrt {33} }}}}{3}}\\
{v = \frac{{\sqrt[3]{{17 + 3\sqrt {33} }}}}{3}}
\end{array}} \right.$
 
$ \Leftrightarrow \left[ {\begin{array}{*{20}{c}}
{y = \frac{{\sqrt[3]{{17 - 3\sqrt {33} }}}}{3} - \frac{2}{{3\sqrt[3]{{17 - 3\sqrt {33} }}}}}\\
{y = \frac{{\sqrt[3]{{17 + 3\sqrt {33} }}}}{3} - \frac{2}{{3\sqrt[3]{{17 + 3\sqrt {33} }}}}}
\end{array}} \right. \Leftrightarrow \left[ {\begin{array}{*{20}{c}}
{t = \frac{{\sqrt[3]{{17 - 3\sqrt {33} }}}}{3} - \frac{2}{{3\sqrt[3]{{17 - 3\sqrt {33} }}}} - \frac{1}{3}}\\
{t = \frac{{\sqrt[3]{{17 + 3\sqrt {33} }}}}{3} - \frac{2}{{3\sqrt[3]{{17 + 3\sqrt {33} }}}} - \frac{1}{3}}
\end{array}} \right.$
 
$ \Leftrightarrow \left[ {\begin{array}{*{20}{c}}
{x = 2\arctan \left( {\frac{{\sqrt[3]{{17 - 3\sqrt {33} }}}}{3} - \frac{2}{{3\sqrt[3]{{17 - 3\sqrt {33} }}}} - \frac{1}{3}} \right) + k\pi }\\
{x = 2\arctan \left( {\frac{{\sqrt[3]{{17 + 3\sqrt {33} }}}}{3} - \frac{2}{{3\sqrt[3]{{17 + 3\sqrt {33} }}}} - \frac{1}{3}} \right) + k\pi }
\end{array}} \right.$

Bài 1 có giải tại: http://www.facebook....&type=1

Ngay từ bước thứ 2, bạn đã sai rồi