Giúp em với! a) Sinx+2cosx +3tan(x/2) =4 b) 1+ 2cos^2(3x/5) =3cos(4x/5) c) 3cos4x - 2cos^2(3x) =1

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