\[\begin{array}{l}
a)\,\,\sin \,x + 2\cos x + 3\tan \,\frac{x}{2} = 4\,\,\left( * \right)\\
DK:\,\,\,\cos \left( {\frac{x}{2}} \right) \ne 0\\
Dat\,\,\tan \,\frac{x}{2} = t \Rightarrow \sin t = \frac{{2t}}{{1 + {t^2}}};\,\,\,\,\,\cos t = \frac{{1 - {t^2}}}{{1 + {t^2}}}\\
\Rightarrow \left( * \right) \Leftrightarrow \frac{{2t}}{{1 + {t^2}}} + 2.\frac{{1 - {t^2}}}{{1 + {t^2}}} + 3t = 4\\
\Leftrightarrow 2t + 2\left( {1 - {t^2}} \right) + 3t\left( {1 + {t^2}} \right) = 4\left( {1 + {t^2}} \right)\\
\Leftrightarrow 2t + 2 - 2t + 3t + 3{t^3} - 4 - 4{t^2} = 0\\
\Leftrightarrow 3{t^3} - 4{t^2} + 3t - 2 = 0\\
\Leftrightarrow \left( {t - 1} \right)\left( {3{t^2} - t + 2} \right) = 0\\
\Leftrightarrow t = 1\\
\Leftrightarrow \tan \frac{x}{2} = 1.\\
b)\,\,1 + 2{\cos ^2}\left( {\frac{{3x}}{5}} \right) = 3\cos \left( {\frac{{4x}}{5}} \right)\\
\Leftrightarrow 1 + \cos \frac{{6x}}{5} + 1 = 3\cos \frac{{4x}}{5}\\
\Leftrightarrow \cos \frac{{6x}}{5} + 2 = 3\cos \frac{{4x}}{5}\\
\Leftrightarrow 4{\cos ^3}\frac{{2x}}{5} - 3\cos \frac{{2x}}{5} + 2 = 6{\cos ^2}\frac{{2x}}{5} - 3\\
\Leftrightarrow 4{\cos ^3}\frac{{2x}}{5} - 3\cos \frac{{2x}}{5} - 6{\cos ^2}\frac{{2x}}{5} + 5 = 0\\
\Leftrightarrow \left( {\cos \frac{{2x}}{5} - 1} \right)\left( {4{{\cos }^2}\frac{{2x}}{5} - 2\cos \frac{{2x}}{5} - 5} \right) = 0
\end{array}\]
Câu c em làm tương tự nhé!!!
